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 $-2c$?

 A: 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.
A: There are a few errors in the question as posed.
First, the quadratic formula is
$$x = \frac{-b \pm \sqrt {b^2-4ac} }{2a}$$
Note the factor of $a$ in the denominator, missing from the OP.  [EDIT -- I see this has been fixed now.]
Second, you write:

In fact, (2) gives solutions for $0x^2+bx+c=0$!

This is not true, and does not even make sense.  The equation $0x^2+bx+c=0$ is equivalent to $bx = -c$, whose only solution is $x=-c/b$ (assuming $b\ne 0$).
In any case, to turn (1) into (2), just multiply by $-b \mp \sqrt{b^2-4ac}$ in both the numerator and the denominator:
$$x = \frac{-b \pm \sqrt {b^2-4ac} }{2a} \cdot \frac{-b \mp \sqrt{b^2-4ac}}{-b \mp \sqrt{b^2-4ac}}$$
The numerator now has the form $(X \pm Y)(Y \pm X)$, which simplifies to just $X^2-Y^2$.  So we have
$$x = \frac{1}{2a} \frac{ \left( -b \right)^2 - \left(b^2-4ac\right)}{-b \mp \sqrt{b^2-4ac}}$$
Cleaning up, the $(-b)^2-b^2$ in the numerator cancels out; the $2a$ in the denominator reduces against the $4ac$ in the numerator, leaving just $2c$; and you can move a negative sign out of the denominator and into the numerator, giving the alternative form of the quadratic formula you wanted.
Finally, you ask

Why is (2) somewhat similar to $\frac{1}{(1)}$ but with $2a$ replaced with $−2c$?

As Ross Millikan says in his answer, if in the original equation $ax^2 + bx + c =0$ we assume that $x=0$ is not a solution (which is equivalent to assuming that $c\ne 0$), then we can divide the entire equation through by $x^2$, obtaining
$$c \left(\frac{1}{x}\right)^2 + b\left(\frac1x\right) + a = 0$$
If we set $u=\frac1x$, then this is $cu^2 + bu + a = 0$, and the quadratic equation tells us
$$u = \frac{ - b \pm \sqrt{b^2 - 4ac} }{2c}$$
Finally we get
$$x = \frac{2c}{-b \pm \sqrt{b^2-4ac}}$$
and you can pull the negative sign out of the denominator into the numerator.  So the reason the equations are so similar is because of a duality in the equation:  interchanging $a$ with $c$ and simultaneously replacing $x$ with $1/x$ changes one quadratic equation into an equivalent one.
A: 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 $-1$
A: Take (2), and rationalize the denominator: 
$$\frac{-2c}{b \pm \sqrt{b^2-4ac}} = \frac{-2c}{b \pm \sqrt{b^2-4ac}}\frac{b \mp \sqrt{b^2-4ac}}{b \mp \sqrt{b^2-4ac}} = $$
$$\frac{-2c(b\mp\sqrt{b^2-4ac})}{b^2-(b^2-4ac)} = \frac{-b \mp \sqrt{b^2-4ac}}{2a}$$
