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 the 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. If a = 0, then the equation becomes linear. We can call it a linear equation. The quadratic equation is of three types namely,

• Standard form
• Factored form
• Vertex form

Generally, there are four different methods to solve the quadratic equation. Those methods are:

• Factoring
• Using square roots
• Completing the squares
• Using quadratic formula

In this, quadratic equation solver page, we will use the quadratic formula to solve the quadratic equation.

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.

The quadratic equation solver uses the quadratic formula to find the roots of the given quadratic equation. The procedure to use the quadratic equation solver is as follows:

Step 1: Enter the coefficients of the quadratic equation “a”, “b” and “c” in the input fields.

Step 2: Now, click the button “Solve the Quadratic Equation” to get the roots.

Step 3: Finally, the discriminant and the roots of the given quadratic equation will be displayed in the output fields.

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

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.

The quadratic equation solution is obtained using the quadratic formula:

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

Normally, we get two solutions, because of a 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

Quadratic Formula 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 \(x=\frac{-1-i\sqrt{47}}{2}\)

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