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First, square both sides, and isolate the absolute value. $$\begin{array}{rl}5|x|+8& =x^2-16\\ 5|x|& =x^2-24\\ |x|& =\frac{x^2-24}{5}.\end{array}$$ Solve for the absolute value and factor. Case 1: $x=\frac{x^2-24}{5}$ Multiplying both sides by $5$ gives us $$5x=x^2-24.$$ Rearranging and factoring, we have $$\begin{array}{rl}{x}^2-5x-24& =0,\\ (x-8)(x+3)& =0.\end{array}$$ Case 2: $x=\frac{-x^2+24}{5}$ As above, we multiply both sides by $5$ to find $$5x=-x^2+24.$$ Rearranging and factoring gives us $$\begin{array}{rl}{x}^2+5x-24& =0,\\ (x+8)(x-3)& =0.\end{array}$$ Combining these cases, we have $x=-8,-3,3,8$. Because our first step of squaring is not reversible, however, we need to check for extraneous solutions. Plug each solution for $x$ back into the original equation to ensure it works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways. Trying $|x|=|3|$, we have $$\begin{array}{rl}\sqrt{5|3|+8}& =\sqrt{3^2-16},\\ \sqrt{15+8}& =\sqrt{9-16},\\ \sqrt{23}& \ne \sqrt{-7}.\end{array}$$ Therefore, $x=3$ and $x=-3$ are extraneous. Checking $|x|=|8|$, we have $$\begin{array}{rl}\sqrt{5|8|+8}& =\sqrt{8^2-16},\\ \sqrt{40+8}& =\sqrt{64-16},\\ \sqrt{48}& =\sqrt{48}.\end{array}$$ The roots of our original equation are $-8$ and $8$ and product is $-8\times 8=\boxed{-64}$.