Proving an alternate quadratic formula 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 $-2c$?


Take (2), and rationalize the denominator: 
$$\frac{-2c}{b \pm \sqrt{b^2-4ac}} = \frac{-2c}{b \pm \sqrt{b^2-4ac}}\frac{b \mp \sqrt{b^2-4ac}}{b \mp \sqrt{b^2-4ac}} = $$
$$\frac{-2c(b\mp\sqrt{b^2-4ac})}{b^2-(b^2-4ac)} = \frac{-b \mp \sqrt{b^2-4ac}}{2a}$$
