What is the sum of the two values of $x$ for which $(x+3)^2 = 121$?
I see that this equation involves a squared term on the left, so I think of using square roots to simplify it. If I take the square root of both sides, I get $x+3 = \pm \sqrt{121}$, where the plus or minus sign accounts for both possible roots. I can simplify the right side further, since $\sqrt{121} = 11$. So I have $x+3 = \pm 11$. Now I can solve for $x$ by subtracting $3$ from both sides. I get $x = \pm 11 - 3$. That means $x = 8$ or $x = -14$. The problem asks for the sum of the two values of $x$, so I just add them together. I get $8 + (-14) = -6$. 