A quadratic equation, which is also called second-degree polynomial equation, holds great importance in algebraic equations. Its standard form or general form is $ax^2+bx+c=0$, where x represents a variable, and a, b, c are real numbers. The formula plays an important role in solving the equations, and students can find online assistance, where expert algebraic tutors assist them to get familiar with quadratic formula and its application.

## History of the Quadratic Formula

Speaking of its history, the biggest contributions made to quadratic equations by the ancient Indian mathematicians. They described a number of methods to solve the quadratic equation. Similarly, Chinese and Babylonian mathematicians played their roles in presenting a method, which involves completing the square, to solve problems related to quadratic equations.

A quadratic equation is a polynomial equation of the second order or degree of equations. In order to solve a quadratic equation, a specific formula is used that is called Quadratic formula. The roots are given by the "quadratic formula" of below:

For the quadratic equation $ax^2 + bx + c = 0$

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

Here, the quadratic formula denotes the both form of real and complex.

x has real values if $b^2-4ac$ > 0.

x has imaginary (complex) values if $b^2-4ac$ < 0.

Standard form of quadratic equation is $ax^2+bx+c=0$ with $a \neq 0$,  where x denotes a variable, and a, b, c are real numbers.

To derive the quadratic formula, students can use the following steps:

 General form of quadratic equation $ax^2 + bx + c = 0$ Step 1: Move constant term to the right side $ax^2 + bx = -c$ Step 2: To make the leading coefficient 1, divide each side by "a". $x^2$ + $\frac{bx}{a}$ = $\frac{-c}{a}$ Step 3: Complete the squareAdd $(\frac{b}{2a})^2$ to both sides $x^2$ + $\frac{bx}{a}$ + $(\frac{b}{2a})^2$ = $\frac{-c}{a}$ + $(\frac{b}{2a})^2$ New form of quadratic equation (x + $\frac{b}{2a}$)$^2$ = $\frac{-c}{a}$ + $\frac{b^2}{4a}$

Step 4:
Solve for x

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

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

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

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

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