Quadratic equation - solving for x Question:
Solve the equation $$(x+2)(x+3)(x+8)(x+12) = 4x^2$$
I tried to solve the equation by expanding the LHS and then equating it to the RHS, but that just doesn't seem to be feasible. I am probably missing a key point here, which would make the question a lot easier. Help please!

I've just got an easier way. Setting $A=x^2+24$ gives you
$$\begin{align}(x+2)(x+3)(x+8)(x+12)=4x^2&\iff (x+2)(x+12)(x+3)(x+8)=4x^2\\&\iff (x^2+14x+24)(x^2+11x+24)=4x^2\\&\iff (A+14x)(A+11x)=4x^2\\&\iff A^2+25xA+150x^2=0\\&\iff (A+10x)(A+15x)=0.\end{align}$$
