Answer
$x\displaystyle \in\left\{-3,\frac{1}{2},1+\sqrt 3, 1-\sqrt 3\right\}$
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, $a_{n}$.
------------------------
$2x^4+x^{3}-17x^{2}-4x+6=0$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2,\pm3,\pm6$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm3, \pm6 ,\pm\frac{1}{2}, \pm\frac{3}{2}$
b. Try for $x=-3:$
$\begin{array}{lllll}
\underline{-3}| & 2 & 1 & -17 & -4& 6\\
& & -6 & 15 & 6&-6\\
& -- & -- & -- & --\\
& 2 & -5 & -2 & 2& |\underline{0}
\end{array}$
$-3$ is a zero,
$(x+3)(2x^3-5x^{2} -2x+2)=0$
To solve for Quadrinomial, Try for $x=\frac{1}{2}$
$\begin{array}{lllll}
\underline{\frac{1}{2}}| & 2 & -5 & -2 & 2\\
& & 1 & -2 & -2\\
& -- & -- & -- & --\\
& 2 & -4 & -4 & |\underline{0}
\end{array}$
$(x+3)(x-\frac{1}{2})(2x^2-4x-4)$
c. To solve for trinomial using the quadratic formula for the quadratic $ax^2+bx+c$, $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}$...
$x=\frac{4\pm\sqrt {(-4)^2-4\times2\times(-4)}}{2\times2}$,
$x=\frac{4\pm\sqrt {48}}{4}$,
$x=\frac{4\pm4\sqrt {3}}{4}$,
$x=1\pm\sqrt {3}$,
$(x-\frac{1}{2})(x+3)(x-1-\sqrt 3)(x-1+\sqrt 3)=0$
$x\displaystyle \in\left\{-3,\frac{1}{2},1+\sqrt 3, 1-\sqrt 3\right\}$
