Is there a way to solve $x^2 + 12y - 12x = 0$ for $x$? I'm doing some statistical analysis (random variate generation for simulation models) and I just ran the inverse transform of a CDF:
$$ F(x) = \begin{cases} (x-4)/4 & \text{for } x \in [2,3] \\ x - (x^2/12) & \text{for } x \in (3,6] \\ 0 & \text{otherwise}
\end{cases} $$
That yieds a couple of equations:
$$ R=(x-4)/4 ~ \text{ for } ~ 2 \leq x \leq 3$$
$$ R=x(1-x)/12 ~ \text{ for } 3 < x  \leq 6 $$
Now, the first one is easy:
$$ 4(R+1)=x ~ \text{ for } -1/2 \leq x \leq -1/4 $$
But the second one is implicit:
$$ (1-(12R/x))1/2=x \text{ for } -2 < x <=-17.5 $$
...backtracking, I rearrange the equation:
\begin{eqnarray*}
R & = & x - (x^2/12) \\
12R & = & 12(x - (x^2/12)) \\
R & = & 12(x - (x^2/12))/12 \\
R & = & (12x - (12x^2/12))/12 \\
R & = &(12x - (x^2))/12 \\
12R & = &(12x - (x^2)) \\
12R & = &12x - (x^2) \\
12R & = &12x - x^2 \\
\end{eqnarray*}
Changing R for y...
$$ x^2 + 12y - 12x= 0 $$
Now, that looks awfully familiar, but I confess I've hit a wall and do not remember what to do from here. 
How can I get an explicit function solving for $x$?

Try the quadratic equation formula:
$$x^2-12x+12y=0\Longrightarrow \Delta:=12^2-4\cdot 1\cdot 12y=144-48y=48(3-y)\Longrightarrow$$
$$x_{1,2}=\frac{12\pm \sqrt{48(3-y)}}{2}=6\pm 2\sqrt{3(3-y)}$$
If you're interested in real roots then it must be that $\,y\le 3\,$ ...
