TITLE: Proving an alternate quadratic formula
QUESTION [8 upvotes]: It is well known that the quadratic formula for $ax^2+bx+c=0$ is given by$$x=\dfrac {-b\pm\sqrt{b^2-4ac}}{2a}\tag1$$
Where $a\ne0$. However, I read somewhere else that given $ax^2+bx+c=0$, we have another solution for $x$ as$$x=\dfrac {-2c}{b\pm\sqrt{b^2-4ac}}\tag2$$
Where $c\ne0$. In fact, $(2)$ gives solutions for $0x^2+bx+c=0$!

Question:


How would you prove $(2)$?
Why is $(2)$ somewhat similar to $\dfrac 1{(1)}$ but with $2a$ replaced with $-2c$?

REPLY [9 votes]: 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.