Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$x=-4$ or $x=4$
Using the definition of the absolute value, we have $$\displaystyle |x^2-16|=0 \\\Longrightarrow x^2-16=0\\ \Longrightarrow x^2=16 \\\Longrightarrow x=\pm\sqrt{16} \\\Longrightarrow x=\pm4$$ Thus, we have: $x=4,$ or $x=-4$