<s>[INST] 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 ${x = -{4\over3}}$.
 [/INST] $12x^2+24x-8x=0\Rightarrow 12x^2+16x=0 \Rightarrow 3x^2+4x=0 \Rightarrow x(3x+4)=0$
So, $x=0$ or $x=- \frac 43$
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