Solving Quadratic Equations in .NET

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Introduction

Hello, and welcome to my small article about Quadratic Equations.

Quadratic Equations

In algebra, a quadratic equation is any equation in the form of ax² + bx + c = 0. In the previous expression, x represents an unknown, and a, b, and c represent known numbers such that a is not equal to 0. The numbers a, b, and c are the coefficients of the equation, and can be distinguished by calling them the quadratic coefficient, the linear coefficient, and the constant or free term.

Our Project

Use Visual Studio to create either a VB.NET or a C# Windows Forms Application and design the form to resemble Figure 1.

Figure 1: Design

Please note that my naming of objects may be different than yours.

Code

Add the following namespaces to your code:

C#

using System;
using System.Windows.Forms;

VB.NET

Imports System.Math

Add the following fields to your form.

C#

      private double sngCoEffA;
      private double sngCoEffB;
      private double sngCoEffC;

      private double sngDiscriminant;

      private double sngRoot1;
      private double sngRoot2;

VB.NET

   Private sngCoEffA As Double
   Private sngCoEffB As Double
   Private sngCoEffC As Double

   Private sngDiscriminant As Double

   Private sngRoot1 As Double
   Private sngRoot2 As Double

In the preceding code, you have created the Coefficient objects, Discriminant objects, as well as the Root objects. Add the following code for the Calculate button.

C#

      private void btnCalculate_Click(object sender, EventArgs e)
      {

         sngCoEffA = float.Parse(txtCoEfficient1.Text.ToString());
         sngCoEffB = float.Parse(txtCoEfficient2.Text.ToString());
         sngCoEffC = float.Parse(txtCoEfficient3.Text.ToString());

         sngDiscriminant = (sngCoEffB * sngCoEffB) - (4 * sngCoEffA
            * sngCoEffC);

         if ((sngDiscriminant > 0))
         {

            sngRoot1 = (-sngCoEffB + Math.Sqrt(sngDiscriminant)) /
               (2 * sngCoEffA);
            sngRoot2 = (-sngCoEffB - Math.Sqrt(sngDiscriminant)) /
               (2 * sngCoEffA);

            txtRoot1.Text = sngRoot1.ToString();
            txtRoot2.Text = sngRoot2.ToString();

         }
         else if ((sngDiscriminant == 0))
         {

            sngRoot1 = (-sngCoEffB + Math.Sqrt(sngDiscriminant)) /
               (2 * sngCoEffA);

            MessageBox.Show("One Root");

            txtRoot1.Text = sngRoot1.ToString();

         }
         else
         {

            MessageBox.Show("No Root");

        }
     }

VB.NET

   Private Sub btnCalculate_Click(ByVal sender As System.Object, _
         ByVal e As System.EventArgs) Handles btnCalculate.Click

      sngCoEffA = txtCoEfficient1.Text
      sngCoEffB = txtCoEfficient2.Text
      sngCoEffC = txtCoEfficient3.Text

      sngDiscriminant = (sngCoEffB * sngCoEffB) - (4 * sngCoEffA _
         * sngCoEffC)

      If (sngDiscriminant > 0) Then

         sngRoot1 = (-sngCoEffB + Math.Sqrt(sngDiscriminant)) / _
            (2 * sngCoEffA)
         sngRoot2 = (-sngCoEffB - Math.Sqrt(sngDiscriminant)) / _
            (2 * sngCoEffA)

         txtRoot1.Text = sngRoot1
         txtRoot2.Text = sngRoot2

      ElseIf (sngDiscriminant = 0) Then

         sngRoot1 = (-sngCoEffB + Math.Sqrt(sngDiscriminant)) / _
            (2 * sngCoEffA)

         MessageBox.Show("One Root")

         txtRoot1.Text = sngRoot1

      Else

         MessageBox.Show("No Root")

      End If

   End Sub

The code for the calculate button works out the Quadratic Equation based on the values that were input. When run, the Form might resemble Figure 2.

Figure 2: Running

Conclusion

Quadratic Equations might seem scary and daunting, especially when complicated formulas are involved. Hopefully, in a follow-up article I can explain a different technique. Until then, happy coding!