Algebraic solution for the intersection point(s) of two parabolas I recently ran through an algebraic solution for the intersection point(s) of two parabolas $ax^2 + bx + c$ and $dx^2 + ex + f$ so that I could write a program that solved for them. The math goes like this:
$$
ax^2 - dx^2 + bx - ex + c - f = 0 \\
x^2(a - d) + x(b - e) = f - c \\
x^2(a - d) + x(b - e) + \frac{(b - e)^2}{4(a - d)} = f - c + \frac{(b - e)^2}{4(a - d)} \\
(x\sqrt{a - d} + \frac{b - e}{2\sqrt{a - d}})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
(a - d)(x + \frac{b - e}{2(a - d)})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
x + \frac{b - e}{2(a - d)} = \sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} \\
x = \pm\sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} - \frac{b - e}{2(a - d)} \\
$$
Then solving for $y$ is as simple as plugging $x$ into one of the equations.
$$
y = ax^2 + bx + c
$$
Is my solution for $x$ and $y$ correct? Is there a better way to solve for the intersection points?

Well, you somehow lost the factor of 4, otherwise seems valid. As you are going to write a program for this, you should also write a solution gfor the case $a=d$ and deal with the case the radical is zero or does not exist.
Actually, you did not have to write all this, as people usually know the solution of quadratic equation (isn't it school material?). For $ax^2 + bx +c = 0$ usually _the discriminant _$D=b^2 - 4ac$ is calculated, and then $x_{1,2}$ are given as $\frac {-b ±\sqrt D}{2a}$. If D=0 then the only sulution $-\frac b {2a}$ exists, and if $D<0$ there are no real solutions (as $\sqrt D$ is not a real number then or does not exist if we only work with real numbers). 
Then just use $a-d,b-e$ and $c-f$ as coefficients in the general solution formula. A separate solution for the case $a=d$.
