The name Quadratic comes from "quad" meaning square as the variable gets squared. It is also called an "Equation of Degree 2".
The standard form of quadratic equation is given as:
$ax^{2}+bx+c=0$, where a, b, c are known and 'x' is the variable which is unknown. Remember 'a' cannot be zero.

As the solutions of some quadratic equations are not rational and we cannot find the solution by factoring. For such equations there is an easy way to solve the solution by using quadratic formula.

The formula giving the roots of quadratic equation $ax^{2}+bx+c=0$ as

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

The equation must be set equal to Zero!
We get two solutions after solving the quadratic equations and they are called roots. The two solutions can be real and distinct also.

The quadratic formula has both real and complex form.

 x has real values if $b^{2} - 4ac > 0$ x has imaginary (complex) values if $b^{2} - 4ac < 0$

The quadratic formula made it appearance around 1100 AD and more or less we know from our high schools. The contributions of the ancient Indian Mathematicians to quadratic equations is highly significant.

Babylonian and Chinese contribution started a method, completing the square to solve basic problems involving area. In 17th century, Rene Descartes developed completing the square method into quadratic formula.

$ax^{2} + bx + c = 0$ where a $\neq$ 0.

Dividing both sides of the equation by a we get,
$x^{2}+\frac{b}{a}x+\frac{c}{a} = 0$

==> $x^{2}+\frac{b}{a}x= -\frac{c}{a}$

Now complete the square by adding $(b/2a)^2$ to both sides

$x^{2}+\frac{b}{a}x+\left(\frac{b}{2a}\right)^{2}= -\frac{c}{a}+\left(\frac{b}{2a}\right)^{2}$

$(x+\frac{b}{2a})^{2}=-\frac{c}{a}+\frac{b^{2}}{4a^{2}}$

$(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}$

Taking square root on both sides

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

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

The above obtained result is known as the Quadratic formula which is usually written as

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

Solved Examples

Question 1: Solve y = $x^{2} +3x -10$ = 0
Solution:

For the given equation we have a = 1, b = 3, c = -10

We know that the formula for quadratic equation is
x=$\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Plugging in the values of a, b, c in the above equation

$\frac{-3\pm\sqrt{3^{2}-4(1)(-10)}}{2(1)}$

$\frac{-3\pm\sqrt{49}}{2}$

$\frac{-3\pm 7}{2}$

Therefore, x = 2 and x = -5

Question 2: Solve y = $x^{2} +12x +20$ = 0
Solution:

For the given equation we have a = 1, b = 12, c = 20

We know that the formula for quadratic equation is
x=$\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Plugging in the values of a, b, c in the above equation

x = $\frac{-12\pm\sqrt{12^{2}-4(1)(20)}}{2(1)}$

x = $\frac{-12\pm\sqrt{64}}{2}$

x = $\frac{-12\pm 8}{2}$

x = -6 $\pm$ 4

Therefore, x = -2 and x= -10