Restrictions on a Function If $f(x) =$
\begin{cases}
x^2-4 &\quad \text{if } x \ge -4, \\
x + 3 &\quad \text{otherwise},
\end{cases}
then for how many values of $x$ is $f(f(x)) = 5$?
I'm not sure how to really start from this question other than bashing values. I need some help on a start, thanks!

First find the values such that $f(x) = 5$:
$x^2-4=5\iff x^2=9 \iff x=\pm 3$ (both are greater than or equal to $-4$)
$x+3=5\iff x = 2$ (but this value is not less than $-4$).
Hence we must have $f(x) = \pm 3$
So now we find the values such that $f(x) = 3$
$x^2-4=3\iff x^2=7 \iff x=\pm \sqrt{7}$ (both are greater than or equal to $-4$)
$x+3=3\iff x = 0$ (but this value is not less than $-4$).
We now find the values such that $f(x) =-3$
$x^2-4=-3\iff x^2=1 \iff x=\pm 1$ (both are greater than or equal to $-4$)
$x+3=-3\iff x = -6$ (this value is less than $-4$).

Hence the values are $\pm\sqrt{7},\pm1$ and $-6$
