C# - Quadratic Equation Solver - ax2 + bx + c = 0
I have given here a C# program to solve any Quadratic Equation. Quadratic equation is a second order of polynomial equation in a single variable.
x = [ -b +/- sqrt(b^2 - 4ac) ] / 2a
We have to find the value of (b*b - 4*a*c).
- When it is greater than Zero, we will get two Real Solutions.
- When it is equal to zero, we will get one Real Solution.
- When it is less than Zero, we will get two Imaginary Solutions.
When there is an imaginary solutions, we have to use the factor i to represent imaginary part as it is a complex number.
Source Code
using System;
using System.Collections.Generic;
using System.Text;
namespace SoftwareAndFinance
{
class Math
{
// quadratic equation is a second order of polynomial equation in a single variable
// x = [ -b +/- sqrt(b^2 - 4ac) ] / 2a
public static void SolveQuadratic(double a, double b, double c)
{
double sqrtpart = b * b - 4 * a * c;
double x, x1, x2, img;
if (sqrtpart > 0)
{
x1 = (-b + System.Math.Sqrt(sqrtpart)) / (2 * a);
x2 = (-b - System.Math.Sqrt(sqrtpart)) / (2 * a);
Console.WriteLine("Two Real Solutions: {0,8:f4} or {1,8:f4}", x1, x2);
}
else if (sqrtpart < 0)
{
sqrtpart = -sqrtpart;
x = -b / (2 * a);
img = System.Math.Sqrt(sqrtpart) / (2 * a);
Console.WriteLine("Two Imaginary Solutions: {0,8:f4} + {1,8:f4} i or {2,8:f4} + {3,8:f4} i", x, img, x, img);
}
else
{
x = (-b + System.Math.Sqrt(sqrtpart)) / (2 * a);
Console.WriteLine("One Real Solution: {0,8:f4}", x);
}
}
static void Main(string[] args)
{
// 6x^2 + 11x - 35 = 0
SolveQuadratic(6, 11, -35);
// 5x^2 + 6x + 1 = 0
SolveQuadratic(5, 6, 1);
// 2x^2 + 4x + 2 = 0
SolveQuadratic(2, 4, 2);
// 5x^2 + 2x + 1 = 0
SolveQuadratic(5, 2, 1);
}
}
}
Click here to get the C# project along with executable
Output
Two Real Solutions: 1.6667 or -3.5000
Two Real Solutions: -0.2000 or -1.0000
One Real Solution: -1.0000
Two Imaginary Solutions: -0.2000 + 0.4000 i or -0.2000 + 0.4000 i
Press any key to continue . . .
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