Proving an alternate quadratic formula

You can write $ax^2+bx+c=0$ as $a+b(\frac 1x)+c(\frac 1x)^2=0,$ solve for $\frac 1x$, then invert it. You can have the minus sign top or bottom as you like by multiplying top and bottom by $-1$

Take (2), and rationalize the denominator:

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

Just for the fun trivia, here are some other "quadratic formulas" $(a>0):$

for $|\frac{b^2}{2a}-2c-1|\le1:$

$$x=-\frac b{2a}\pm\frac1{\sqrt a}\cos\left(\frac12\arccos\left(\frac{b^2}{2a}-2c-1\right)\right)$$

for $\frac{b^2}{2a}-2c-1>1:$

$$x=-\frac b{2a}\pm\frac1{\sqrt a}\cosh\left(\frac12\operatorname{arccosh}\left(\frac{b^2}{2a}-2c-1\right)\right)$$

You can play around with some graphs here if you want.