problem: The quadratic $x^2-6x+66$ can be written in the form $(x+b)^2+c$, where $b$ and $c$ are constants. What is $b+c$?
solution: We complete the square.
  
  We have $(x-3)^2 = x^2 - 6x + 9$, and so
  
  \begin{align*}
  x^2-6x+66 &= (x-3)^2 - 9 + 66 \\
  &= (x-3)^2 + 57.
  \end{align*}Therefore, $b=-3$ and $c=57$, which gives us $b+c = \boxed{54}$.