How to Solve Quadratic Equation

There are many techniques for how to solve quadratic equations. For example, we can solve a quadratic equation by factoring or directly using the quadratic formula. In this post, we will learn about the same.

How to solve quadratic equation

Solving Quadratic Equations using Factoring 

To solve a quadratic equation using the factoring method, we will follow the below steps given below:

  1. First, write down the given equation in the standard form, (i.e., ax2 + bx + c = 0) in which one side contains the variable and constants, and the other side will be zero.
  2. Next, factor the expression of the non-zero side.
  3. Equate each of the factors of the previous step with 0. This is because a product of non-zero factors is zero if and only if one or more of the factors is 0.
  4. Solve each equation obtained in Step 3.


Solve the equation x2 -5x + 6 = 0.


First, factor x2 -5x + 6. See that

x2 -5x + 6

= x2 -3x -2x + 6

= x(x-3) -2(x-3)

= (x-2)(x-3)

So the solution of x2 -5x + 6 = 0 are given by

x-2=0 or x-3=0

⇒ x=2, 3

Thus, x=2,3 are the solutions of x2-5x+6=0.

Steps to Solve a Quadratic Equation

The general method of solving a quadratic equation is described below. The steps are as follows:

  1. First, express the given equation in the standard form: ax2 + bx + c = 0, where a, b, c are constants and x is the variable.
  2. Next, check whether ax2 + bx + c can be factored or not. If it is, then follow the steps given above.
  3. If ax2 + bx + c cannot be factored, then we proceed as follows:

ax2 + bx + c =0

⇒ $a \left(x^2 + \dfrac{b}{a}x + \dfrac{c}{a} \right) =0$

⇒ $x^2 + \dfrac{b}{a}x + \dfrac{c}{a} =0$ as a is non-zero; otherwise, the given equation won’t be quadratic.

⇒ x2 + 2⋅x ⋅$\dfrac{b}{2a}$ + $(\dfrac{b}{2a})^2$ – $(\dfrac{b}{2a})^2$ + $\dfrac{c}{a}$ =0

⇒ [x2 + 2⋅x ⋅$\dfrac{b}{2a}$ + $(\dfrac{b}{2a})^2$] – $\dfrac{b^2}{4a^2}$ + $\dfrac{c}{a}$ =0

⇒ $\left(x + \dfrac{b}{2a} \right)^2$ – $\dfrac{b^2-4ac}{4a^2}$ =0 using the formula a2+2ab+b2=(a+b)2

⇒ $\left(x + \dfrac{b}{2a} \right)^2$ = $\dfrac{b^2-4ac}{4a^2}$

Taking square roots on both sides, we get that

x + $\dfrac{b}{2a}$ = ± $\dfrac{\sqrt{b^2-4ac}}{2a}$

⇒ x = – $\dfrac{b}{2a}$ ± $\dfrac{\sqrt{b^2-4ac}}{2a}$

⇒ x = $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ …(∗)

So the solutions of the given equation ax2 + bx + c = 0 are x= [-b+√(b2-4ac)]/2a and [-b-√(b2-4ac)]/2a. This method of solving quadratic equations is known as Shreedhara Acharya’s formula, named after the ancient Indian mathematician Shreedhara Acharya who derived it. ♣

Let us now understand the above Shreedhara Acharya’s formula for solving a quadratic equation.


Solve the equation x2 + x + 1 = 0.


Comparing the given equation x2 + x + 1 = 0 with ax2 + bx + c = 0, we get that

a=1, b=1, c=1.

By the above Shreedhara Acharya’s formula (∗), the solutions of x2 + x + 1 = 0 will be

x = $\dfrac{-1 \pm \sqrt{1^2-4\cdot 1\cdot 1}}{2\cdot 1}$

= $\dfrac{-1 \pm \sqrt{-3}}{2}$

= $\dfrac{-1 \pm \sqrt{3}i}{2}$

So the solutions of x2 + x + 1 = 0 are $\dfrac{-1 + \sqrt{3}i}{2}$ and $\dfrac{-1 – \sqrt{3}i}{2}$, where i=√-1 is an imaginary complex number.


How to solve a linear equation

Solve x2+25=0


Q1: How to solve a quadratic equation?

Answer: A quadratic equation can be solved by either factoring or by the quadratic formula (Shreedhara Acharya’s formula).

Q2: What is Shreedhara Acharya’s formula for solving quadratic equations?

Answer: The solutions of the equation ax2 + bx + c = 0 are x= [-b+√(b2-4ac)]/2a and [-b-√(b2-4ac)]/2a. This formula is known as Shreedhara Acharya’s formula.

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