 A quadratic equation solver is a free step by step solver for solving the quadratic equation to find the values of the variable. With the help of this solver, we can find the roots of quadratic equation given by, ax2 + bx + c = 0, where the variable x has two roots. The solution is obtained using the quadratic formula; where a, b and c are the real numbers and a ≠ 0.

## How does the Quadratic Equation Solver Work?

A quadratic equation is nothing but a polynomial of degree 2. The roots of polynomials give the solution of the equation. Here we have to solve an equation in the form of ax2 + bx + c = 0.

Enter the values of a, b and c in the solver given below to solve any given quadratic equation.

Quadratic Equation : ax2 + bx + c = 0

Enter the value of a :
Enter the value of b :
Enter the value of c :

Discriminant (D):

x1:
x2:

where x1 and x2 are root 1 and root 2.

## Steps to Solve Quadratic Equation

The input for the quadratic equation solver is of the form

ax2 + bx + c = 0

Where a is not zero, a ≠ 0

If the value of a is zero, then the equation is not a quadratic equation.

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Normally, we get two solutions, because of plus or minus symbol “±”. You need to do both the addition and subtraction operation.

The part of an equation “ b2-4ac “ is called the “discriminant” and it produces the different types of possible solutions. Some of the possible solutions are

• Case 1: When a discriminant part is positive, you get two real solutions
• Case 2: When a discriminant part is zero, it gives only one solution
• Case 3: When a discriminant part is negative, you get complex solutions

Quadratic solver level helps the students of class 10 to clearly know about the different cases involved in the discriminant producing different solutions. Here are some of the quadratic equation examples

• Case 1 : b2 – 4ac > 0

Example 1: Consider an example x2 – 3x – 10 = 0

Given data : a =1, b = -3 and c = -10

b2 – 4ac = (-3)2– 4 (1)(-10)

= 9 +40 = 49

b2 – 4ac= 49 >0

Therefore, we get two real solutions

The general quadratic formula is given as;

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ $x=\frac{-(-3)\pm \sqrt{(-3)^{2}-4(1)(-10)}}{2(1)}$ $x=\frac{3\pm \sqrt{9+40}}{2}$ $x=\frac{3\pm \sqrt{49}}{2}$ $x=\frac{3\pm 7}{2}$

x= 10/2 , -4/2

x= 5, -2

Therefore, the solutions are 5 and -2

• Case 2 : b2 – 4ac = 0

Example 2: Consider an example 9x2 +12x + 4 = 0

Given data : a =9, b = 12 and c = 4

b2 – 4ac = (12)2– 4 (9)(4)

= 144 – 144= 0

b2 – 4ac= 0

Therefore, we get only one distinct solution

The general quadratic formula is given as

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ $x=\frac{-(12)\pm \sqrt{(12)^{2}-4(9)(4)}}{2(9)}$ $x=\frac{-12\pm \sqrt{144-144}}{18}$ $x=\frac{-12\pm \sqrt{0}}{18}$ $x=\frac{-12}{18}$

x= -6/9 = -2/3

x= -2/3

Therefore, the solution is -2 / 3

• Case 3 : b2 – 4ac < 0

Example 3: Consider an example x2 + x + 12= 0

Given data : a =1, b = 1 and c = 12

b2 – 4ac = (1)2– 4 (1)(12)

= 1 – 48 = -47

b2 – 4ac= -47 < 0

Therefore, we get complex solutions

The general quadratic formula is given as

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ $x=\frac{-(1)\pm \sqrt{(1)^{2}-4(1)(12)}}{2(1)}$ $x=\frac{-1\pm \sqrt{1-48}}{2}$ $x=\frac{-1\pm \sqrt{-47}}{2}$ $x=\frac{-1+i\sqrt{47}}{2}$ and $x=\frac{-1-i\sqrt{47}}{2}$

Therefore, the solutions are

$x=\frac{-1+i\sqrt{47}}{2}$ and $x=\frac{-1-i\sqrt{47}}{2}$

For more information about quadratic equations and other related topics in mathematics, register with BYJU’S – The Learning App and watch interactive videos.