jmc

$x$ is not in the domain of $f$ if the denominator is zero. Since both absolute values are nonnegative, both must be zero for the denominator to be zero. So

$$\begin{array}{rl}0=x^2+3x-4=(x+4)(x-1)& \Rightarrow x=-4\text{ or }x=1\\ 0=x^2+9x+20=(x+4)(x+5)& \Rightarrow x=-4\text{ or }x=-5\end{array}$$

The only value of $x$ which makes both absolute values zero is $x=\boxed{-4}$.