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.

$
\begin{split}
f(x) &= -3(x^2-5x/3 - 1/3)\\
   &= -3( x^2 - 5x/3 + (5/6)^2 - (5/6)^2 - 1/3)\\
   &= -3( (x - 5/6)^2 - 25/36 - 12/36) \\
   &= -3(x - 5/6)^2 -3(- 25/36 - 12/36) \\
   &= -3(x - 5/6)^2 + 37/12.
\end{split}$
