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Having trouble in finding roots of a cubic equation with C#

Recently my friends and I decided to make a multi-functional calculator to automatically find
the roots of quadratic equations.
It wasn't that challenging after all, so we decided to go onto the next level, to make a calculator for a cubic equation.
(ax^3 + bx^2 + cx + d)
However, we stumbled across some trivial problems, and no matter how hard we tried, the results are still the same.
we are beginners in terms of coding, so we're not sure if we are actually making some stupid mistakes here
but at least we want to learn something from others.
Basically, we have tried a lot of combinations of different cubic equations, or even re-coding the whole thing. The problem is that the results we yield are always wrong, but only for the real parts of the second root and third root.
For a better understanding, we have tried 9x^3 + 8x^2 + 7x + 6, as an example.
The correct answer, according to a cubic equation calculator website, is
(The website)
First root = -0.87285
Second root = -0.00802 + 0.87391 i
Third root = -0.00802 - 0.87391 i
However, our result to this equation is :
First root = -0.87285
Second root = -0.2963 + 0.87391 i
Third root = -0.2963 + -0.87391 i
It is apparently noticeable that only parts of the second and third roots are wrong.
We have tried finding similar threads to help, but those are a little bit too difficult for us to understand or those problems are not the same as ours.
We look forward to finding the solution and the causes to this problem.
We have separated the formulas for finding the roots of a cubic equation into 5 parts.
(rootp1-rootp5)
The formulas are coded according to the formulas that can be found in the Chinese version of Wikipedia's page of Cubic equation.
(The formulas)
We have also rounded the real parts of the roots to 5 digits.
(Some code lines might be redundant, but as I have mentioned, we are new to coding)
Code (C#) :
using System;
using System.Numerics;
namespace ComplexNumbers
{
public class ComplexNo
{
public static void Main()
{
Console.WriteLine("Cubic Equation Calculator :");
Console.Write("Insert first coefficient : ");
double a = Convert.ToDouble(Console.ReadLine());
Console.Write("Insert second coefficient : ");
double b = Convert.ToDouble(Console.ReadLine());
Console.Write("Insert third coefficient : ");
double c = Convert.ToDouble(Console.ReadLine());
Console.Write("Insert constant : ");
double d = Convert.ToDouble(Console.ReadLine());
Console.WriteLine(" ");
Console.WriteLine("Solve for " + a + "x" + "^3 " + b + "x^2 " + c + "x " + d + " :");
double rootp1 = -(b / (3 * a));
double rootp2 = (b * c / (6 * Math.Pow(a, 2))) - (Math.Pow(b, 3) / (27 * Math.Pow(a, 3))) - (d / (2 * a));
double rootp3 = (c / (3 * a)) - (Math.Pow(b, 2) / (9 * Math.Pow(a, 2)));
Complex root1 = rootp1 + Math.Cbrt(rootp2 + Math.Sqrt(Math.Pow(rootp2, 2) + Math.Pow(rootp3, 3))) +
Math.Cbrt(rootp2 - Math.Sqrt(Math.Pow(rootp2, 2) + Math.Pow(rootp3, 3)));
Complex rootp4 = new Complex(-1 / 2, Math.Sqrt(3) / 2);
Complex rootp5 = new Complex(-1 / 2, -(Math.Sqrt(3) / 2));
Complex root2 = rootp1 + (rootp4 * Math.Cbrt(rootp2 + Math.Sqrt(Math.Pow(rootp2, 2) + Math.Pow(rootp3, 3)))) +
(rootp5 * Math.Cbrt(rootp2 - Math.Sqrt(Math.Pow(rootp2, 2) + Math.Pow(rootp3, 3))));
Complex root3 = rootp1 + (rootp5 * Math.Cbrt(rootp2 + Math.Sqrt(Math.Pow(rootp2, 2) + Math.Pow(rootp3, 3)))) +
(rootp4 * Math.Cbrt(rootp2 - Math.Sqrt(Math.Pow(rootp2, 2) + Math.Pow(rootp3, 3))));
Console.WriteLine(" ");
Console.WriteLine("Results :");
Console.WriteLine("First Root :");
string root1rp = Convert.ToString(Math.Round(root1.Real, 5));
string root1ip = Convert.ToString(Math.Round(root1.Imaginary, 5));
Console.WriteLine(root1rp + " + " + root1ip + "i");
Console.WriteLine("Second Root :");
string root2rp = Convert.ToString(Math.Round(root2.Real, 5));
string root2ip = Convert.ToString(Math.Round(root2.Imaginary, 5));
Console.WriteLine(root2rp + " + " + root2ip + "i");
Console.WriteLine("Third Root :");
string root3rp = Convert.ToString(Math.Round(root3.Real, 5));
string root3ip = Convert.ToString(Math.Round(root3.Imaginary, 5));
Console.WriteLine(root3rp + " + " + root3ip + "i");
Console.ReadLine();
}
}
}
(Sorry for making this thread so long and my bad grammar)

The problem is with this line Complex rootp4 = new Complex(-1 / 2, Math.Sqrt(3) / 2);. -1/2 uses integer division and evaluates to 0.
This code will work.
Console.WriteLine(" ");
Console.WriteLine("Solve for " + a + "x" + "^3 " + b + "x^2 " + c + "x " + d + " :");
double minusBover3a = -(b / (3 * a));
double rootp2 = (b * c / (6 * Math.Pow(a, 2))) - (Math.Pow(b, 3) / (27 * Math.Pow(a, 3))) - (d / (2 * a));
double rootp3 = (c / (3 * a)) - (Math.Pow(b, 2) / (9 * Math.Pow(a, 2)));
double bigCubeRootPlus = Math.Cbrt(rootp2 + Math.Sqrt(Math.Pow(rootp2, 2) + Math.Pow(rootp3, 3)));
double bigCubeRootMinus = Math.Cbrt(rootp2 - Math.Sqrt(Math.Pow(rootp2, 2) + Math.Pow(rootp3, 3)));
// ** THIS IS THE PROBLEM. "-1/2" uses integer division, so this complex has 0 for real part
Complex complexPlus = new Complex(-1.0 / 2, Math.Sqrt(3) / 2);
Complex complexMinus = new Complex(-1.0 / 2, -(Math.Sqrt(3) / 2));
Complex root1 = minusBover3a + bigCubeRootPlus + bigCubeRootMinus;
Complex root2 = minusBover3a + complexPlus * bigCubeRootPlus + complexMinus * bigCubeRootMinus;
Complex root3 = minusBover3a + complexMinus * bigCubeRootPlus + complexPlus * bigCubeRootMinus;

I admit I haven't tried your code, but you may have a casting issue in the math in your original calculations. Try this:
double rootp2 = ((double)b * (double)c / (6D * Math.Pow(a, 2D))) - (Math.Pow(b, 3D) / (27D * Math.Pow(a, 3D))) - ((double)d / (2D * (double)a));
If that makes a difference, you'd have to propagate a similar change (cast all variables as (double) and inline constants as double with D) through the other calculations.

The following is an alternative to your code:
// from https://www.daniweb.com/programming/software-development/
// code/454493/solving-the-cubic-equation-using-the-complex-struct
// algorithm described in
// https://en.wikipedia.org/wiki/Cubic_equation#General_cubic_formula
const int NRoots = 3;
double SquareRootof3 = Math.Sqrt(3);
// the 3 cubic roots of 1
var CubicUnity = new List<Complex>(NRoots)
{ new Complex(1, 0),
new Complex(-0.5, -SquareRootof3 / 2.0),
new Complex(-0.5, SquareRootof3 / 2.0) };
// intermediate calculations
double DELTA = 18 * a * b * c * d
- 4 * b * b * b * d
+ b * b * c * c
- 4 * a * c * c * c
- 27 * a * a * d * d;
double DELTA0 = b * b - 3 * a * c;
double DELTA1 = 2 * b * b * b
- 9 * a * b * c
+ 27 * a * a * d;
Complex DELTA2 = -27 * a * a * DELTA;
Complex C = Complex.Pow((DELTA1 + Complex.Pow(DELTA2, 0.5)) / 2, 1 / 3.0);
for (int i = 0; i < NRoots; i++)
{
Complex M = CubicUnity[i] * C;
Complex Root = -1.0 / (3 * a) * (b + M + DELTA0 / M);
Console.WriteLine();
Console.WriteLine($"Root {i+1}:");
Console.WriteLine($"Real {Root.Real:0.#####}");
Console.WriteLine($"Imaginary {Root.Imaginary:0.#####}i");
}

How do I show cents when I am writing a code to count total money, the code below is only showing a whole number (such as $1)

I am trying to write a code in C# that prints out the total of dollars and cents I have. I have the first part done, but the end of the code won't show how much cents there is when I run the whole code, it only shows a whole number (such as $1), when I need the code to show something like $1.56. What am I missing in the program in order to show this?
Console.WriteLine("Enter number of quarters: ");
int quarters = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Enter number of dimes: ");
int dimes = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Enter number of nickels: ");
int nickels = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Enter number of pennies: ");
int pennies = Convert.ToInt32(Console.ReadLine());
double dollars = (int)(((quarters * 0.25) + (dimes * 0.10) + (nickels * 0.05) + (pennies * 0.01)) % 1 * 100);
Console.WriteLine("Your total is $" + dollars);
When I typed in 4 quarters, 0 dimes, 0 nickels, and 1 penny, it shows the result as $1 not $1.01

You are forcing the result as int, so the decimals are cut
double dollars = (int)(((quarters * 0.25) + (dimes * 0.10) + (nickels * 0.05) + (pennies * 0.01)) % 1 * 100);
Do not force to int, on the contrary, force to double
double dollars = ((quarters * 0.25) + (dimes * 0.10) + (nickels * 0.05) + (pennies * 0.01)) % 1 * 100;
But, since you are handling money, I would recommend to handle everything with decimal
decimal dollars = ((quarters * 0.25) + (dimes * 0.10) + (nickels * 0.05) + (pennies * 0.01)) % 1 * 100;
Better explained in dotnet perls
Decimal accurately stores numeric data. The .NET Framework offers this
type for programs where rounding errors are harmful. Decimal stores
large and small numbers with many digits after the decimal place.
EDIT
When I put in decimal dollars instead of double dollars, I then get
the error, "Cannot implicitly convert type 'double' to 'decimal'. An
explicit conversion exists (are you missing a cast?)
It is becuse you need all the operation of the same type, you can do
int quarters =1;
int dimes = 1;
int nickels = 1;
int pennies = 1;
decimal dollars = ((quarters * 0.25m) + (dimes * 0.10m) + (nickels * 0.05m) + (pennies * 0.01m)) % 1 * 100;
Console.WriteLine("Your total is $" + dollars);
Note the use of a m after the number, this is to tell the compiler that the type is decimal or Money, not double.

How to show solutions for quadratic formula

So I wrote a Quadratic formula program in C#, how do I take the quadratic formula program and modify it so that the program correctly displays the number of solutions.
if there are two solutions,
(x - x1)(x - x2) = 0
if there is only one solution,
(x - x0)^2 = 0
if there are no solutions,
No Solution.
This is the program, if someone could show the solution to this for me that would be wonderful, I'm really stuck on how to do it.
using System;
namespace quadraticequation
{
class MainClass
{
public static void Main(string[] args)
{
Console.WriteLine("Enter a number for a"); //ask the user for information
double a = double.Parse(Console.ReadLine()); //Gets a from the user
Console.WriteLine("Enter a number for b"); //asks the user for information
double b = double.Parse(Console.ReadLine()); //Gets b from the user
Console.WriteLine("Enter a number for c"); //asks the user for information
double c = double.Parse(Console.ReadLine()); //Gets c from the user
//double.Parse --> is used to convert a number or string to a double.
//Console.ReadLine() --> is used to take the input from the user.
//We call a function here
Quadratic(a, b, c);
}
//We need to create a new function
public static void Quadratic(double a, double b, double c)
{
double deltaRoot = Math.Sqrt(b * b - 4 * a * c); //Math.Sqrt takes the square root of the number
if (deltaRoot >= 0) // we use an if statement here to handle information
{
double x1 = (-b + deltaRoot) / 2 * a; //We write the information for x1 here
double x2 = (-b - deltaRoot) / 2 * a; //We write the information for x2 here
Console.WriteLine("x1 = " + x1 + " x2 = " + x2); //we use this to write the roots
}
else // we use an else statement so that we dont return an error when there are no roots
{
Console.WriteLine("There are no roots");
}
}
}
}

I think you have to review your second degree formula solution-skills. You write:
double deltaRoot = Math.Sqrt(b * b - 4 * a * c);
But the test is actually whether b2-4×a×c is larger than or equal to zero: indeed that is actually why we check it: because we cannot take the square root of a negative number (yeah there exist complex numbers that can take the square root of a negative number, but let's ignore that for now).
So the solution is to write it like:
public static void Quadratic(double a, double b, double c) {
double delta = b*b-4*a*c; //only delta
if (delta > 0) {
double deltaRoot = Math.Sqrt(delta);
double x1 = (-b + deltaRoot) / (2 * a); //We write the information for x1 here
double x2 = (-b - deltaRoot) / (2 * a); //We write the information for x2 here
Console.WriteLine("x1 = " + x1 + " x2 = " + x2); //we use this to write the roots
} else if(delta == 0) {
double x1 = -b/(2*a);
Console.WriteLine("x1 = " + x1); //we use this to write the roots
} else {
Console.WriteLine("There are no roots");
}
}
You also have to write (-b + deltaRoot) / (2*a) (with (2*a)), otherwise you will multiply (-b + deltaRoot) / 2 with a instead.
A final note is that equality comparisons with floating points is very tricky so delta == 0 will often fail since the result can be something 1e-20-ish, which is simply an error when performing floating point arithmetic. So it is perhaps better to use a small range of values.
This gives:
csharp> MainClass.Quadratic(1,1,1);
There are no roots
csharp> MainClass.Quadratic(1,1,0);
x1 = 0 x2 = -1
csharp> MainClass.Quadratic(1,0,0);
x1 = 0

Calculation results in NaN [closed]

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Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
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Improve this question
I was trying to calculate the result of a second degree equation using C#, but i get the result NaN. Can someone check my code to locate the error ?
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace desafio2_2
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Please insert the values that multiply x², x and the independent term respectively: ");
float a = float.Parse(Console.ReadLine());
float b = float.Parse(Console.ReadLine());
float c = float.Parse(Console.ReadLine());
double bhaskarap1 = (Math.Pow(b, 2)) + (- 4 * a * c);
double raiz1 = (-b + Math.Sqrt(bhaskarap1)) / (2 * a);
double raiz2= (-b - Math.Sqrt(bhaskarap1)) / (2 * a);
Console.WriteLine(raiz1);
Console.WriteLine(raiz2);
}
}
}

It may be possible that the quadratic equation has no real roots. In this case the discriminant will be negative.
Math.Sqrt(x) returns NaN when when x is negative. You should test the discriminant for negative number before you call the square root function.
static void Main(string[] args)
{
Console.WriteLine("Please insert the values that multiply x², x and the independent term respectively: ");
float a = float.Parse(Console.ReadLine());
float b = float.Parse(Console.ReadLine());
float c = float.Parse(Console.ReadLine());
double bhaskarap1 = (Math.Pow(b, 2)) + (- 4 * a * c);
if (bhaskarap1 < 0)
{
Console.WriteLine("There are no real solutions.");
return;
}
double raiz1 = (-b + Math.Sqrt(bhaskarap1)) / (2 * a);
double raiz2= (-b - Math.Sqrt(bhaskarap1)) / (2 * a);
Console.WriteLine(raiz1);
Console.WriteLine(raiz2);
}

argument of sqrt method must be positive.

I would check for NaN using a ternary operator like I did below to handle the case of a negative value for sqrt.
float a = 4.0f, b = 7.0f, c = -3.0f;
double my_var = ((Math.Pow(b, 2)) + (-4 * a * c));
Console.WriteLine(my_var);
Console.WriteLine(((-b + Math.Sqrt(-1 * my_var)) / (2 * a)));
double temp = (-b + Math.Sqrt(my_var)) / (2 * a);
double raiz1 = temp.CompareTo(Double.NaN) < 1 ? 0.00 : temp;
temp = (-b + Math.Sqrt(-1 * my_var)) / (2 * a);
double raiz2 = temp.CompareTo(Double.NaN) < 1 ? 0.00 : temp;
Console.WriteLine($"Raiz1: {raiz1}\tRaiz2: {raiz2}");
Output:
97
NaN
Raiz1: 0.356107225224513 Raiz2: 0

The roots of the second degree equations are complex numbers in general case, e.g.
x**2 + 1 = 0
That's why I suggest using Complex, not float:
...
using System.Numerics;
...
// float.Parse - if you want to allow just real coefficients
Complex a = float.Parse(Console.ReadLine());
Complex b = float.Parse(Console.ReadLine());
Complex c = float.Parse(Console.ReadLine());
Complex bhaskarap1 = b * b - 4 * a * c;
Complex raiz1 = (-b + Complex.Sqrt(bhaskarap1)) / (2 * a);
Complex raiz2 = (-b - Complex.Sqrt(bhaskarap1)) / (2 * a);
Console.WriteLine(raiz1);
Console.WriteLine(raiz2);

x++ operator for playing gif animations: How do it work together in this example? [duplicate]

I have next code
int a,b,c;
b=1;
c=36;
a=b%c;
What does "%" operator mean?

It is the modulo (or modulus) operator:
The modulus operator (%) computes the remainder after dividing its first operand by its second.
For example:
class Program
{
static void Main()
{
Console.WriteLine(5 % 2); // int
Console.WriteLine(-5 % 2); // int
Console.WriteLine(5.0 % 2.2); // double
Console.WriteLine(5.0m % 2.2m); // decimal
Console.WriteLine(-5.2 % 2.0); // double
}
}
Sample output:
1
-1
0.6
0.6
-1.2
Note that the result of the % operator is equal to x – (x / y) * y and that if y is zero, a DivideByZeroException is thrown.
If x and y are non-integer values x % y is computed as x – n * y, where n is the largest possible integer that is less than or equal to x / y (more details in the C# 4.0 Specification in section 7.8.3 Remainder operator).
For further details and examples you might want to have a look at the corresponding Wikipedia article:
Modulo operation (on Wikipedia)

% is the remainder operator in many C-inspired languages.
3 % 2 == 1
789 % 10 = 9
It's a bit tricky with negative numbers. In e.g. Java and C#, the result has the same sign as the dividend:
-1 % 2 == -1
In e.g. C++ this is implementation defined.
See also
Wikipedia/Modulo operation
References
MSDN/C# Language Reference/% operator

That is the Modulo operator. It will give you the remainder of a division operation.

It's the modulus operator. That is, 2 % 2 == 0, 4 % 4 % 2 == 0 (2, 4 are divisible by 2 with 0 remainder), 5 % 2 == 1 (2 goes into 5 with 1 as remainder.)

It is the modulo operator. i.e. it the remainder after division 1 % 36 == 1 (0 remainder 1)

That is the modulo operator, which finds the remainder of division of one number by another.
So in this case a will be the remainder of b divided by c.

It's is modulus, but you example is not a good use of it. It gives you the remainder when two integers are divided.
e.g. a = 7 % 3 will return 1, becuase 7 divided by 3 is 2 with 1 left over.

It is modulus operator
using System;
class Test
{
static void Main()
{
int a = 2;
int b = 6;
int c = 12;
int d = 5;
Console.WriteLine(b % a);
Console.WriteLine(c % d);
Console.Read();
}
}
Output:
0
2

is basic operator available in almost every language and generally known as modulo operator.
it gives remainder as result.

Okay well I did know this till just trying on a calculator and playing around so basically:
5 % 2.2 = 0.6 is like saying on a calculator 5/2.2 = 2.27 then you multiply that .27 times the 2.27 and you round and you get 0.6. Hope this helps, it helped me =]

Nobody here has provided any examples of exactly how an equation can return different results, such as comparing 37/6 to 37%6, and before some of you get upset thinking that you did, pause for a moment and think about it for a minute, according to Dirk Vollmar in here the int x % int y parses as (x - (x / y) * y) which seems fairly straightforward at first glance, but not all Math is performed in the same order.
Since not every equation has it's proper brackets, some Schools will teach that the Equation is to be parsed as ((x - (x / y)) * y) whilst other Schools teach (x - ((x / y) * y)).
Now I experimented with my example (37/6 & 37%6) and figured out which selection was intended (it's (x - ((x / y) * y))), and I even displayed a nicely built if Loop (even though I forgot every End of Line Semicolon) to simulate the Divide Equation at the most Fundamental Scale, as that was in fact my point, the Equation is similar, yet Fundamentally different.
Here's what I can remember from my deleted Post (this took me around an Hour to Type from my Phone, indents are Double Spaced)
using System;
class Test
{
static void Main()
{
float exact0 = (37 / 6); //6.1666e∞
float exact1 = (37 % 6); //1
float exact2 = (37 - (37 / 6) * 6); //0
float exact3 = ((37 - (37 / 6)) * 6); //0
float exact4 = (37 - ((37 / 6) * 6)); //185
int a = 37;
int b = 6;
int c = 0;
int d = a;
int e = b;
string Answer0 = "";
string Answer1 = "";
string Answer2 = "";
string Answer0Alt = "";
string Answer1Alt = "";
string Answer2Alt = "";
Console.WriteLine("37/6: " + exact0);
Console.WriteLine("37%6: " + exact1);
Console.WriteLine("(37 - (37 / 6) * 6): " + exact2);
Console.WriteLine("((37 - (37 / 6)) * 6): " + exact3);
Console.WriteLine("(37 - ((37 / 6) * 6)): " + exact4);
Console.WriteLine("a: " + a + ", b: " + b + ", c: " + c + ", d: " + d + ", e: " + e);
Console.WriteLine("Answer0: " + Answer0);
Console.WriteLine("Answer0Alt: " + Answer0Alt);
Console.WriteLine("Answer1: " + Answer1);
Console.WriteLine("Answer0Alt: " + Answer1Alt);
Console.WriteLine("Answer2: " + Answer2);
Console.WriteLine("Answer2Alt: " + Answer2Alt);
Console.WriteLine("Init Complete, starting Math...");
Loop
{
if (a !< b) {
a - b;
c +1;}
if else (a = b) {
a - b;
c +1;}
else
{
String Answer0 = c + "." + a; //6.1
//this is = to 37/6 in the fact that it equals 6.1 ((6*6=36)+1=37) or 6 remainder 1,
//which according to my Calculator App is technically correct once you Round Down the .666e∞
//which has been stated as the default behavior of the C# / Operand
//for completion sake I'll include the alternative answer for Round Up also
String Answer0Alt = c + "." + (a + 1); //6.2
Console.WriteLine("Division Complete, Continuing...");
Break
}
}
string Answer1 = ((d - (Answer0)) * e); //185.4
string Answer1Alt = ((d - (Answer0Alt​)) * e); // 184.8
string Answer2 = (d - ((Answer0) * e)); //0.4
string Answer2Alt = (d - ((Answer0Alt​) * e)); //-0.2
Console.WriteLine("Math Complete, Summarizing...");
Console.WriteLine("37/6: " + exact0);
Console.WriteLine("37%6: " + exact1);
Console.WriteLine("(37 - (37 / 6) * 6): " + exact2);
Console.WriteLine("((37 - (37 / 6)) * 6): " + exact3);
Console.WriteLine("(37 - ((37 / 6) * 6)): " + exact4);
Console.WriteLine("Answer0: " + Answer0);
Console.WriteLine("Answer0Alt: " + Answer0Alt);
Console.WriteLine("Answer1: " + Answer1);
Console.WriteLine("Answer0Alt: " + Answer1Alt);
Console.WriteLine("Answer2: " + Answer2);
Console.WriteLine("Answer2Alt: " + Answer2Alt);
Console.Read();
}
}
This also CLEARLY demonstrated how an outcome can be different for the exact same Equation.