Answer
$x\in \{-\sqrt5i, \sqrt5i, 2\}$
Work Step by Step
See The Rational Zero Theorem:
... If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients,
then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=x^4-4x^3+9x^2-20x+20,$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2,\pm4,\pm5$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm4,\pm5$
b. Try for $x=2:$
$\begin{array}{lllll}
\underline{2}| &1& -4 & 9 & -20 & 20\\
& & 2&-4 & 10&-20\\
& -- & -- & -- & --\\
& 1&-2 & 5&-10 & |\underline{0}
\end{array}$
$f(x)=(x-2)(x^3-2x^2+5x-10),$
$(x^3-2x^2+5x-10),$
$x^2(x-2)+5(x-2),$
$(x^2+5)(x-2)^2,$
$f(x)=(x^2+5)(x-2)^2=(x-\sqrt 5i)(x+\sqrt 5i)(x-2)^2$
$x\in \{-\sqrt5i, \sqrt5i, 2\}$
