quadratic equation in c# programming code example

Example 1: quadratic C#

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);

        }

    }

}

Example 2: Program to find number of solutions in Quadratic Equation c#

static void checkSolution(int a, int b, int c) 
    { 
  
        // If the expression is greater than 0,  
        // then 2 solutions 
        if (((b * b) - (4 * a * c)) > 0) 
            System.out.println("2 solutions"); 
  
        // If the expression is equal 0, then 2 solutions 
        else if (((b * b) - (4 * a * c)) == 0) 
            System.out.println("1 solution"); 
  
        // Else no solutions 
        else
            System.out.println("No solutions"); 
    }