Answer
$x\in\{-2, -1, 2, 5\}$
Work Step by Step
$x^4-4x^3-9x^2+16x+20=0$,
$\begin{array}{lllll}
\underline{5}| & 1 & -4 & -9 & 16& 20\\
& & 5 & 5 & -20& -20\\
\hline & & & & \\
& 1 & 1 & -4 & -4& 0
\end{array}$
$(x^4-4x^3-9x^2+16x+20) \div(x-5)=x^3+x^2-4x-4$.
we can factor out the quadrinomial just using distributive property as follows,
$(x^3+x^2-4x-4)(x-5)$
$x^2(x+1)-4(x+1)$,
$(x^2-4)(x+1)$,
$(x-2)(x+2)(x+1)(x-5)$,
$x\in\{-2, -1, 2, 5\}$
