Answer
$x\displaystyle \in\left\{-1, \frac{-1- \sqrt {61}}{6}, \frac{-1+ \sqrt {61}}{6}, 3\right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f$ with integer coefficients,
then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=3x^4-5x^{3}-16x^{2}+7x+15$
Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 3, \pm 5, \pm15$
$q:\qquad \pm 1, \pm 3$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 3, \pm5, \pm15 ,\pm\frac{1}{3}, \pm \frac{5}{3}$
Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}|& 3 & -5 & -16 & 7 & 15\\
& & -3 & 8& 8& -15\\
& -- & -- & -- & --\\
& 3 & -8 & -8& 15 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(3x^3-8x^2-8x+15)$
Try for $x=3$:
$\begin{array}{lllll}
\underline{3}|& 3 & -8 & -8 & 15\\
& & 9& 3& -15\\
& -- & -- & -- & --\\
& 3 & 1& -5 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x+1)(x-3)(3x^2+x-5)$
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, $3x^2+x-5$, $x=\frac{-1\pm \sqrt {1^2-4 \times 3\times (-5)}}{2\times 3}=\frac{-1\pm \sqrt {61}}{6}$.
Therefore, the real zeros of the equation are:
$x\displaystyle \in\left\{-1, \frac{-1- \sqrt {61}}{6}, \frac{-1+ \sqrt {61}}{6}, 3\right\}$
