Answer
$x\in \{-3, -1, -\frac{1}{2}, 1\}$
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, $a_{n}$.
------------------------
$f(x)=2x^4+7x^3+x^2-7x-3,$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 3,$
$q:\qquad \pm 1, \pm2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 3, \pm \frac{1}{2}, \pm\frac{3}{2}$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| &2& 7 & 1 & -7 & -3\\
& & 2 & 9 & 10&3\\
& -- & -- & -- & --\\
& 2 & 9 & 10&3 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(2x^3+9x^2+10x+3)$
Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}| & 2 & 9 & 10 & 3\\
& & -2 & -7&-3\\
& -- & -- & -- & --\\
& 2 & 7&3 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x-1)(x+1)(2x^2+7x+3)$
$2x^2+7x+3,$
$2x^2+x+6x+3,$
$x(2x+1)+3(2x+1),$
$(x+3)(2x+1),$
$f(x)=(x-1)(x+1)(x+3)(2x+1)$
$x\in \{-3, -1, -\frac{1}{2}, 1\}$
