Given 2 x=3+5i
$\Rightarrow \quad x=\frac{3+5 i}{2}$
Consider $x^{3}=\frac{27+125 i^{3}+225 i^{2}+135 i}{8}$
$=\frac{27-125 i-225+135 i}{8} \quad\left(\begin{array}{r}\because i^{2}=-1 \\ i^{3}=-i\end{array}\right)$
$=\frac{-198+10 i}{8}=\frac{-99+5 i}{4}$
and $x^{2}=\frac{9+25 i^{2}+30 i}{4}$
$=\frac{9-25+30 i}{4}=\frac{-8+15 i}{2}$
Now, Consider $2 x^{3}+2 x^{2}-7 x+72$
$=\left(\frac{-99+5 i}{2}\right)+(-8+15 i)-\frac{7(3+5 i)}{2}+72$
$=-\frac{99}{2}+\frac{5 i}{2}-8+15 i-\frac{21}{2}-\frac{35}{2} i+72$
$=\left(-\frac{99}{2}-8-\frac{21}{2}+72\right)+\left(\frac{5}{2}+15-\frac{35}{2}\right) i$
$=\frac{-99-16-21+144}{2}=\frac{8}{2}=4$