algebra precalculus - 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?

Answer

we know that

$$(ay - b)^2 = a^2y^2 - 2ayb + b^2 \space \space \space \space (1)$$

we have

$$6 + 12y - 36y^2 = -(36y^2 - 12y - 6)$$

we need to factor

$$36y^2 - 12y - 6$$

from (1) let $a^2 = 36 \Rightarrow a = \pm 6, 2ab = 12 \Rightarrow b = \pm 1$

thus

$$36y^2 - 12y + 1 - 7 = (6y - 1)^2 \ or \ (-6y +1)^2 - 7 \Rightarrow 6 + 12y - 36y^2 = 7 - (6y-1)^2$$