Answer
$x\in \{-3, -\frac{1}{\sqrt2}i, \frac{1}{\sqrt2}i, 2\}$
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, $a_{n}$.
------------------------
$f(x)=2x^4+2x^3-11x^2+x-6,$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2,\pm3,$
$q:\qquad \pm 1,\pm2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm3,\pm\frac{1}{2},\pm\frac{3}{2}$
b. Try for $x=2:$
$\begin{array}{lllll}
\underline{2}| &2& 2 & -11 & 1& -6\\
& & 4&12 & 2&6\\
& -- & -- & -- & --\\
& 2&6 & 1&3 & |\underline{0}
\end{array}$
$f(x)=(x-2)(2x^3+6x^2+x+3),$
$(2x^3+6x^2+x+3),$
$2x^2(x+3)+1(x+3),$
$(2x^2+1)(x+3),$
$f(x)=(x-2)(2x^2+1)(x+3)$
$x\in \{-3, -\frac{1}{\sqrt2}i, \frac{1}{\sqrt2}i, 2\}$
