Answer
$\quad x\displaystyle \in\left\{\frac{5}{2}\right\}$
Work Step by Step
Factoring in pairs is not obvious here, so we try to find rational zeros of
$f(x)=2x^{3}-3x^{2}-3x-5$
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 5}{\pm 1,\pm 2}$
Testing with synthetic division, ... we eventually try $x-\displaystyle \frac{5}{2}$
$\left.\begin{array}{l}
5/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 &-3 &-3 &-5 \\\hline
&5 &5 & 5 \\\hline
2& 2 &2 & |\ \ 0 \end{array}$
$f(x)=(x-\displaystyle \frac{5}{2})2( x^{2}+x+1)$
applying the quadratic formula,
$b^{2}-4ac=1-4=-3$
the discriminant is negative,
$ x^{2}+x+1=0\quad$ has no real solutions.
$ f(x)=0\quad$when$\quad x\displaystyle \in\left\{\frac{5}{2}\right\}$
