If $f(y)=ax^2+bx+c$, does this imply that $x=\frac{-b \pm \sqrt{b^2-4a[c-f(y)]}}{2a}$?

The equation $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ is known as the quadratic formula and is the solution to the quadratic equation $ax^2+bx+c=0$.

Sometimes I encounter equations such as $x=y^2-y$. Is it correct to say that since we can hold $x$ as a constant, we can write the equation as $y^2-y-x=0$ which has the form $ay^2+by+c=0$, where $a=1$,$b=-1$ and $c=-x$, thus giving us the following formula for $y=f(x)$: $$y=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-x)}}{2(1)}=\frac{1}{2} \pm \sqrt{1+4x}$$ ?

-
Yes. What is the source of you being unsure?, maybe we can discuss that. –  Lord Soth Apr 22 '13 at 19:35
(Of course, $a$ should be non-zero). –  Lord Soth Apr 22 '13 at 19:36
The fraction on the last part isn't correct. If $x=0$ then the solutions should be 0,1 not $\frac{1}{2}$ and $-\frac{1}{2}$ –  JB King Apr 22 '13 at 19:44
I was skeptical because usually the quadratic formula is used for constants. I didn't know it could be used for variables, but I suspected that it would work for variables because variables are just 'varying constants' in the sense that we can build a graph of infinite relations between $x$ and $y$ by using the equation infinite times. –  raindrop Apr 23 '13 at 15:51
If you wanna see it intuitively for sure, use the quadratic formula and then write the original in factored form. Not only can you distribute to check that you have not changed the equations at all, but also you should be able to see very clearly that the only solutions are the ones you originally found with the quadratic formula. –  Ovi Apr 24 '13 at 6:08