Answer
$x\in \{-2, \frac{1}{2}\}$
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, $a_{n}$.
------------------------
$f(x)=4x^3+4x^2-7x+2,$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2,$
$q:\qquad \pm 1,\pm2,\pm4 $
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm\frac{1}{2},\pm\frac{1}{4}$
b. Try for $x=\frac{1}{2}:$
$\begin{array}{lllll}
\underline{\frac{1}{2}}| & 4 & 4 & -7 & 2\\
& & 2 & 3&-2\\
& -- & -- & -- & --\\
& 4 & 6&-4 & |\underline{0}
\end{array}$
$f(x)=(2x-1)(4x^2+6x-4),$
$4x^2-2x+8x-4,$
$2x(2x-1)+4(2x-1),$
$(2x+4)(2x-1),$
$f(x)=2(2x-1)^2(x+2)$
$x\in \{-2, \frac{1}{2}\}$
