First things first. We have to get that into standard form. [tex]-8x^2+4x+12[/tex] is what I'm seeing. Hope I did not misinterpret. In order to complete the square, we need to do 2 things. First let's set the quadratic equal to 0, then move over the 12 by subtraction. [tex]-8x^2+4x=-12[/tex]. The rule for completing the square is that the leading coefficient HAS to be a +1. Ours is a -8, so we have to factor it out. When we do that, we will get [tex]-8(x^2- \frac{1}{2}x)=-12 [/tex]. Now we are ready to complete the square. Take half the linear term, square it, and add it to both sides. Our linear term is 1/2. 1/2 divided by 2 is 1/4. If we square 1/4 we get 1/16. So we add it in. HOWEVER, we have that -8 hanging around out front on the left, refusing to be ignored. If we add in 1/16 into the parenthesis, we have to multiply it by -8 and add whatever that product is onto the right side. -8*1/16= -1/2. This is what all that looks like: [tex]-8(x^2- \frac{1}{2}x+ \frac{1}{16})=-12- \frac{1}{2} [/tex]. In this process we have created a perfect square binomial on the left, which is this: [tex]-8(x- \frac{1}{4})^2=- \frac{25}{2} [/tex]. If I'm not mistaken, that's the form you need it in.
