Standard Form for a Parabola What is the standard form for the following problem? I already know that it is a horizontal parabola. I just can't seem to be able to change it into the standard format.

$8y² +96y-12x+240 = 0$

I have gotten it to $x = 2/3 (y + 6)² - 4$, but that's not right.

Edit: How do you find the focus of this parabola?
$12x = 8(y^2+12y+36-6) = 8(y+6)^2 - 48 \to x -(-4)= \dfrac{2(y-(-6))^2}{3}$. The focus of this parabola lies on the axis of symmetry which is the line $y = -6$. We have: $F = (-4+p,-6)$, with $p = \dfrac{1}{4a} = \dfrac{1}{4\cdot \frac{2}{3}} = \dfrac{3}{8} \to F = (\frac{-29}{8},-6)$.
Edit: I think the problem might be that it is not a vertical parabola.  The $y$ is squared, so it must be a sideways parabola. (Also called a horizontal parabola).