How do you factor this using complete the square? $6+12y-36y^2$ I'm so embarrassed that I'm stuck on this simple algebra problem that is embedded in an integral, but I honestly don't understand how this is factored into $a^2-u^2$
Here are my exact steps:
$6+12y-36y^2$ can be rearranged this way: $6+(12y-36y^2)$ and I know I can factor out a -1 and have it in this form: $6-(-12y+36y^2)$
This is the part where I get really lost.  According to everything I read, I take the $b$ term, which is $-12y$ and divide it by $2$ and then square that term.  I get: $6-(36-6y+36y^2)$
The form it should look like, however, is $7-(6y-1)^2$
Can you please help me to understand what I'm doing wrong?

Here is the first step:
$$6+12y-36y^2 = -\left((6y)^2-2\cdot (6y)-6\right)$$
A complete square would be
$$\left(6y-1\right)^2 = (6y)^2-2\cdot (6y) + 1.$$
Ignoring the overall minus for a moment and adding and subtracting 1 to the r.h.s. of the first equation above, we get (using $a^2-b^2=(a-b)(a+b)$ formula:
$$(6y)^2-2\cdot (6y) - 6 = \left(6y-1\right)^2 - 7 = \left(6y-1 - \sqrt{7}\right) \left(6y-1+\sqrt{7}\right)$$
And finally, restoring the overall minus gets us the answer:
$$6+12y-36y^2 = - \left(6y-1 - \sqrt{7}\right) \left(6y-1+\sqrt{7}\right).$$
