Completing the square help The textbook gives this equation:
${12x^2 + 24x -8x = 0}$ with an answer of ${x = 0}$ or ${x = -{4\over3}}$
But I suspect it should be ${12x^2 + 24x -8 = 0}$
So in order to solve this, I would first isolate the x terms on one side of the equation by adding 8 to both sides:
${12x^2 + 24x = 8}$
I would then divide both sides by the coefficient of the ${x^2}$ or 12 in this case which gives:
${x^2 + 2x = {8\over12}}$
I then divide the coefficient of x by 2 and square the result and add it to both sides
${x^2 + 2x + 1 = {8\over12}}$
=> ${(x + 1)^2 = {8\over12}}$
=> ${x + 1 = \pm \sqrt{8\over12}}$
=> ${x + 1 = \pm \sqrt{2\over3}}$
=> ${x = - 1\pm \sqrt{2\over3}}$
I've taken a wrong turn somewhere, I'm not sure how to get to ${x = 0}$ or ${x = -{4\over3}}$.

No, if you consider $12x^2 + 24x = 8$, then considering $x=0$ you get $0=8$, which is clearly wrong.
Considering $12x^2 + 24x = 8x$ this is the same as $12x^2 + 16x = 0$, and from here:
$$ x(12x + 16) = 0$$
Which is fulfilled if $x=0$ or $12 x = -16 \implies x = - \frac{4}{3}$
