Originally Posted by
HallsofIvy I can't speak for the actual history but it was probably by using "completing the square" since that is the easiest way to derive it:
With the equation $\displaystyle ax^2+ bx+ c= 0$ we can write it as $\displaystyle a\left(x^2+ \frac{b}{a}x\right)+ c= 0$. We can complete the square by adding (and subtracting)$\displaystyle \left(\frac{b}{2a}\right)^2$ inside the parentheses. $\displaystyle a\left(x^2+ \frac{b}{a}x+ \frac{b^2}{4a^2}- \frac{b^2}{4a}\right)+ c$$\displaystyle = a\left(x^2+ \frac{b}{a}x+ \frac{b^2}{4a^2}\right)- \frac{b^2}{4a}+ c$$\displaystyle = \left(x+ \frac{b}{2a}\right)^2+ c- \frac{b^2}{4a^2}= 0$.
$\displaystyle \left(x+ \frac{b}{2a}\right)^2= \frac{b^2}{4a^2}- c= \frac{b^2- 4ac}{4a^2}$.
Taking the square root of both sides, $\displaystyle x+ \frac{b}{2a}= \pm\frac{\sqrt{b^2- 4ac}}{2a}$.
So $\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$.