Calculating the point of intersection of two parabolas
Well it seems as though I SIGNIFICANTLY overcomplicated the equation for some reason. I just did the equation it a second time and I got something much nicer:

For any two parabolas \$ ax^2 + bx + c = 0\$ and \$dx^2 + ex + f = 0\$, we get:

\$
ax^2 - dx^2 + bx - ex + c - f = 0 \\
x^2(a - d) + x(b - e) = f - c
\$

Now it's just a matter of completing the square.

\$
x^2(a - d) + x(b - e) + \frac{(b - e)^2}{4(a - d)} = f - c + \frac{(b - e)^2}{4(a - d)} \\
(x\sqrt{a - d} + \frac{b - e}{2\sqrt{a - d}})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
(a - d)(x + \frac{b - e}{2(a - d)})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
x + \frac{b - e}{2(a - d)} = \sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} \\
x = \sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} - \frac{b - e}{2(a - d)} \\
\$

I don't know how I ended up with the monster I originally posted, but this seems much more reasonable.