Answer
$x\displaystyle \in\left\{ \frac{1- \sqrt {3}}{2}, \frac{1+ \sqrt {3}}{2},\frac{1}{2}\right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ 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, $a_{n}$.
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$f(x)=4x^{3}-6x^{2}+1$
Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1,$
$q:\qquad \pm 1, \pm2, \pm4$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}$
Try for $x=\frac{1}{2}:$
$\begin{array}{lllll}
\underline{\frac{1}{2}}|& 4& -6 & 0 & 1\\
& & 2& -2& -1\\
& -- & -- & -- & --\\
& 4 & -4& -2 & |\underline{0}
\end{array}$
$\frac{1}{2}$ is a zero,
$f(x)=(x-\frac{1}{2})(4x^2-4x-2)$
Solving for the trinomial, using the quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$.
in this case, $4^2-4x-2$, $x=\frac{4\pm \sqrt {-4^2-4 \times 4\times (-2)}}{2\times 4}=\frac{4\pm 4\sqrt {3}}{8}=\frac{1\pm\sqrt 3}{2}$.
Therefore, the real zeros of the equations are:
$x\displaystyle \in\left\{ \frac{1- \sqrt {3}}{2}, \frac{1+ \sqrt {3}}{2},\frac{1}{2}\right\}$
