derivation of quadratic formula

Suppose $A,B,C$ are real numbers, with $A\ne 0$, and suppose

$$A{x}^2+Bx+C=0.$$

Since $A$ is nonzero, we can divide by $A$ and obtain the equation

$$x^2+bx+c=0,$$

where $b=\frac{B}{A}$ and $c=\frac{C}{A}$.This equation can be written as

$$x^2+bx+\frac{b^2}{4}-\frac{b^2}{4}+c=0,$$

so completing the square, i.e., applying the identity ${(p+q)}^2=p^2+2pq+q^2$, yields

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

Then, taking the square root of both sides, and solving for $x$, we obtainthe solution formula

$x$$=$$-{\displaystyle \frac{b}{2}}\pm \sqrt{{\displaystyle \frac{b^2}{4}}-c}$

$=$${\displaystyle \frac{B}{2A}}\pm \sqrt{{\displaystyle \frac{B^2}{4A^2}}-{\displaystyle \frac{C}{A}}}$

$=$${\displaystyle \frac{-B\pm \sqrt{B^2-4AC}}{2A}},$

and the derivation is completed.

A slightly less intuitive but more aesthetically pleasing approach to this derivation can be achieved by multiplying both sides of the equation

$a{x}^2+bx+c=0$

by $4a$, resulting in the equation

$4a^2{x}^2+4abx+b^2=b^2-4ac,$

in which the left-hand side can be expressed as ${(2ax+b)}^2$. From here, the proof is identical.