Answer
$x\displaystyle \in\left\{-1, -\frac{1}{2}, -3-\sqrt {10}, -3+\sqrt {10}\right\}$
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, $a_{n}$.
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$f(x)=2x^4+15x^{3}+17x^{2}+3x-1$
Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, $
$q:\qquad \pm 1, \pm2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm \frac{1}{2}$
Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}|& 2 & 15 & 17 & 3 & -1\\
& & -2 & -13& -4& 1\\
& -- & -- & -- & --\\
& 2 & 13 & 4& -1 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(2x^3+13x^2+4x-1)$
Try for $x=-\frac{1}{2}$:
$\begin{array}{lllll}
\underline{-\frac{1}{2}}|& 2 & 13 & 4 & -1\\
& & -1& -6& 1\\
& -- & -- & -- & --\\
& 2 & 12& -2 & |\underline{0}
\end{array}$
$-\frac{1}{2}$ is a zero,
$f(x)=(x+1)(x+\frac{1}{2})(2x^2+12x-2)$
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, $2x^2+12x-2$, $x=\frac{-12\pm \sqrt {12^2-4 \times 2\times (-2)}}{2\times 2}=\frac{-12\pm 4\sqrt {10}}{4}=-3\pm\sqrt {10}$.
Therefore, the real zeros of the equations are:
$x\displaystyle \in\left\{-1, -\frac{1}{2}, -3-\sqrt {10}, -3+\sqrt {10}\right\}$
