Answer
$x\displaystyle \in\left\{ -2, \frac{-1- \sqrt {5}}{2}, \frac{-1+ \sqrt {5}}{2},1\right\}$
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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$f(x)=x^4+2x^{3}-2x^{2}-3x+2$
Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2$
Try for $x=1:$
$\begin{array}{lllll}
\underline{1}|& 1 & 2 & -2 & -3 & 2\\
& & 1 & 3& 1& -2\\
& -- & -- & -- & --\\
& 1 & 3 & 1& -2 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(x^3+3x^2+x-2)$
Try for $x=-2$:
$\begin{array}{lllll}
\underline{-2}|& 1 & 3 & 1 & -2\\
& & -2& -2& 2\\
& -- & -- & -- & --\\
& 1 & 1& -1 & |\underline{0}
\end{array}$
$-2$ is a zero,
$f(x)=(x+2)(x-1)(x^2+x-1)$
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, $x^2+x-1$, $x=\frac{-1\pm \sqrt {1^2-4 \times 1\times (-1)}}{2\times 1}=\frac{-1\pm \sqrt {5}}{2}$.
Therefore, the real zeros of the equations are:
$x\displaystyle \in\left\{ -2, \frac{-1- \sqrt {5}}{2}, \frac{-1+ \sqrt {5}}{2},1\right\}$
