Answer
$x\in \{ -3,\frac{1}{4},2 \}$
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, $a_{n}$.
------------------------
$4x^3+3x^2-25x+6=0,$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm2, \pm3, \pm 6, ,$
$q:\qquad \pm 1,\pm2,\pm4$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm2, \pm3, \pm 6, \pm\frac{1}{2}, \pm\frac{3}{2},$
b. Try for $x=2:$
$\begin{array}{lllll}
\underline{2}| &4& 3 & -25 & 6\\
& & 8&22 & -6\\
& -- & -- & -- & --\\
& 4&11& -3 & |\underline{0}
\end{array}$
Try for $x=-3:$
$\begin{array}{lllll}
\underline{-3}| &4 & 11 & -3\\
& & -12 & 3\\
& -- & -- & -- & --\\
& 4&-1 & |\underline{0}
\end{array}$
$(x-2)(x+3)(4x-1),$
$x\in \{ -3,\frac{1}{4},2 \}$
