Completing the square with negative x coefficients I know how to complete the square with positive $x$ coefficients but how do you complete the square with negative $x$ coefficients?
For example:
\begin{align*}
f(x) & = x^2 + 6x + 11 \\
 & = (x^2 + 6x) + 11 \\
 & = (x^2 + 6x + \mathbf{9}) + 11 - \mathbf{9} \\
 & = (x+3)^2 + 2.
\end{align*}
For positive you would add 9 inside the parenthesis and subtract 9 outside. However, I get the wrong answer when I do it with a negative coefficient. Do you do the same thing when there is a negative coefficient in front of $x^2$ or is it the other way around? (subtract inside parenthesis and add outside of parenthesis).
Here is an example of a negative coefficient:
$$f(x) = -3x^2 + 5x + 1.$$
I tried to solve this and entered in my answer online but it was wrong.

$$(2ax+b)^2 = 4a^2x^2 + 4abx + b^2$$
So, multiplying by $4a$, or in this case $-4a$, will avoid fractions until the last step.
\begin{align}
   f(x) &= -3x^2 + 5x + 1 \\
   -12f(x) &= 36x^2 -60x - 12 \\
   -12f(x) &= 36x^2 - 60x + 25 - 25 - 12 \\
   -12f(x) &= (6x-5)^2 - 37 & \left(b = \frac{4ab}{4a} = \dfrac{-60}{12}=-5   \right)\\
   f(x) &= -\dfrac{1}{12}(6x-5)^2 + \dfrac{37}{12}
\end{align}
