Quadratic equation program is not giving values

Question

Could you tell me what's wrong with my code for solving a quadratic equation? Don't be cringed out by my work, because I'm still very new to the program.

class Program
{
    static void Main(string[] args)
    {
        double a, b, c, x1, x2, x, D;
        String A;
        String B;
        String C;
        Console.Write("a=");
        A = Console.ReadLine(); 
        Console.Write("b=");
        B = Console.ReadLine();
        Console.Write("c=");
        C = Console.ReadLine();
        a = Convert.ToDouble(A);
        b = Convert.ToDouble(B);
        c = Convert.ToDouble(C);
        D =(b * b - 4 * a * c);
        if (D > 0)
        {
            x1 = (-b + Math.Sqrt(D)) / (2 * a);
            x2 = (-b - Math.Sqrt(D)) / (2 * a);
            Console.WriteLine("x1=" + x1);
            Console.WriteLine("x2=" + x2);
        }
        else
            if (D < 0)
            {
                D = -D;
                x1 = (-b + Math.Sqrt(D)) / (2 * a);
                x2 = (-b - Math.Sqrt(D)) / (2 * a);
            }
            else
            {
                x = (-b / (2 * a));
                Console.WriteLine("x=" + x);
            }
        Console.ReadKey();

    }
}

It doesn't display any values when it is run.

Can you explain what is not working with your code so far. – strongbutgood Oct 01 '16 at 18:52 — strongbutgood, Oct 01 '16 at 18:52
Well, when i run it, i don't get any values. – Адам Никић Oct 01 '16 at 19:14 — Адам Никић, Oct 01 '16 at 19:14

Clock · Answer 1 · 2016-10-01T19:09:45.000

Try something like

class Program
{
    static void Main(string[] args)
    {
        double a, b, c, x1, x2, x, D;
        String A;
        String B;
        String C;
        Console.Write("a=");
        A = Console.ReadLine();
        Console.Write("b=");
        B = Console.ReadLine();
        Console.Write("c=");
        C = Console.ReadLine();
        a = Convert.ToDouble(A);
        b = Convert.ToDouble(B);
        c = Convert.ToDouble(C);

        if (a != 0)
        {
            D = (b * b - 4 * a * c);

            if (D > 0)
            {
                x1 = (-b + Math.Sqrt(D)) / (2 * a);
                x2 = (-b - Math.Sqrt(D)) / (2 * a);
                Console.WriteLine("x1=" + x1);
                Console.WriteLine("x2=" + x2);
            }
            else
            {
                if (D < 0)
                {
                    D = -D;
                    //x1 = (-b + Math.Sqrt(D)) / (2 * a);
                    //x2 = (-b - Math.Sqrt(D)) / (2 * a);

                    string complexX1 = ((-b) / (2 * a)).ToString() + "+" + (Math.Sqrt(D) / 2 * a).ToString() + "i";
                    string complexX2 = ((-b) / (2 * a)).ToString() + "-" + (Math.Sqrt(D) / 2 * a).ToString() + "i";

                    Console.WriteLine("complex x1=" + complexX1);
                    Console.WriteLine("complex x2=" + complexX2);
                }
                else
                {
                    x = (-b / (2 * a));
                    Console.WriteLine("x=" + x);
                }
            }
        }
        else
        {
            if (b != 0)
            {
                x = -c / b;

                Console.WriteLine("Grade 1 equation");
                Console.WriteLine("x=" + x);
            }
            else
            {
                Console.WriteLine("No equation.");
            }
        }

        Console.ReadKey();
    }
}

It seems that you were not writing to console in all the cases.

And also the way how the complex roots were treated was not entirely correct since the "i" was missing so that the real and imaginary part of the number were missing. Now the result should be fine also for complex roots.

I updated the code so now also cases when a is 0 (the equation is not quadric) and b is 0 (there is no equation) are handle.

May I know who down voted my answer and why ? The answer is really a nice solution for a quadric equation. — Clock, Oct 01 '16 at 18:50
You're welcome ! Please do not forgate to mark the answer :D. — Clock, Oct 01 '16 at 19:15
So i tried the code, and it does give right numbers, but they're negative, though the shouldn't be. — Адам Никић, Oct 01 '16 at 19:18
Oh, it was my fault. I typed in 9 instead of -9. The equation was a^2-9a+20 — Адам Никић, Oct 01 '16 at 19:22
When working with *complex* values (roots in the case) why not suing `Complex` structure which is specially desinged for that purpose? https://msdn.microsoft.com/en-us/library/system.numerics.complex(v=vs.110).aspx — Dmitry Bychenko, Oct 01 '16 at 20:04
I think that for a begginer is better to keep things simpler. — Clock, Oct 01 '16 at 20:21

Dmitry Bychenko · Answer 2 · 2016-10-01T21:17:42.277

Do not cram everything into a single Main method, decompose your implementation! In your very case, let's extract a method:

   // IEnumerable<Double>: quadratic equation can have 0..2 distinct real roots  
   private static IEnumerable<Double> SolveQuadratic(double a, double b, double c) {
     double D = b * b - 4 * a * c;

     if (D < 0)
       yield break; // No real solutions (they are both complex ones)
     else if (D == 0)
       yield return -b / (2 * a); // One distinct root    
     else {
       // Two distinct solutions
       yield return (-b - Math.Sqrt(D)) / (2 * a); 
       yield return (-b + Math.Sqrt(D)) / (2 * a); 
     }
   }

Then use it:

   ...
   a = Convert.ToDouble(A);
   b = Convert.ToDouble(B);
   c = Convert.ToDouble(C);

   var result = SolveQuadratic(a, b, c)
     .Select((x, i) => $"x{i + 1} = {x}");  

   Console.Write(String.Join(Environment.NewLine, result));

   Console.ReadKey();

Quadratic equation program is not giving values

2 Answers2