What is the improved quadratic formula to solve quadratic equations?

Hint: There is only one quadratic formula and that is the Dharacharya formula to solve quadratic equations. Here in this question, we will derive this formula. It is the only formula that helps us to find the roots of a quadratic equation.

Complete step by step answer:
The quadratic formula to find the roots of quadratic equation $a{x^2} + bx + c = 0$ , where $a \ne 0$ is given by:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
The plus and minus sign represents that quadratic formula will give two roots, one root corresponding to the plus sign and the other root corresponding to the minus sign.
 i.e. ${x_1} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$ and ${x_2} = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$.
Now we will derive this formula.
Consider the quadratic equation $a{x^2} + bx + c = 0$ where a, b, c are real numbers and $a \ne 0$.
$a{x^2} + bx + c = 0$
Transpose c to RHS
$ \Rightarrow a{x^2} + bx = - c$
Now, multiply both sides by $4a$
$ \Rightarrow 4{a^2}{x^2} + 4abx = - 4ac$
Now add ${b^2}$both sides
$ \Rightarrow 4{a^2}{x^2} + 4abx + {b^2} = {b^2} - 4ac$
Now, we can factorize
$ \Rightarrow {\left( {2ax + b} \right)^2} = {b^2} - 4ac$
Now, taking root both sides
$ \Rightarrow \left( {2ax + b} \right) = \pm \sqrt {{b^2} - 4ac} $
Transpose b to right hand side
$ \Rightarrow 2ax = - b \pm \sqrt {{b^2} - 4ac} $
Now, divide both sides by $2a$
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Therefore, the roots of the quadratic equation are ${x_1} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$ and ${x_2} = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$.

Note:
The expression ${b^2} - 4ac$ is called the discriminant of the quadratic equation. It can be positive, negative and zero also. If it is positive then we get a distinct and real solution, if it is negative then we get no real solution and if it is zero then we have real and equal solutions.