Solving $\log _2(x-4) + \log _2(x+2) = 4$ Here is how I have worked it out so far:
$\log _2(x-4)+\log(x+2)=4$
$\log _2((x-4)(x+2)) = 4$
$(x-4)(x+2)=2^4$
$(x-4)(x+2)=16$
How do I proceed from here?
$x^2+2x-8 = 16$
$x^2+2x = 24$
$x(x+2) = 24$ Which I know is not the right answer
$x^2+2x-24 = 0$ Can't factor this

The quadratic formula:
$$ x = \frac{ -b \pm \sqrt{b^2 - 4 a c} }{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-24) }}{2 \times 1} $$
4 + 96 in the square root.
