Answer
$(\displaystyle \frac{f}{g})(x)= x^2-3x-13$
Work Step by Step
$(\displaystyle \frac{f}{g})(x)=\frac{2x^{3}-9x^{2}-17x+39}{2x-3}$
$ \begin{array}{llllllllll}
& & x^{2} & -3x & -13 & & & \color{red}{ \small{Quotient}} & & \\
& & -- & -- & -- & -- & -- & -- & & \\
2x-3 & ) & 2x^{3} & -9x^{2} & -17x & +39 & & & & \\
& & 2x^{3} & -3x^{2} & & & & \color{red}{\leftarrow \small{ x^{2}(2x-3) } } & & \\
& & -- & -- & & & & \color{red}{\leftarrow \small{ subtract } } & & \\
& & & -6x^{2} & -17x & +39 & & & & \\
& & & -6x^{2} & +9x & & & \color{red}{\leftarrow \small{-3x(2x-3) } } & & \\
& & & -- & -- & & & \color{red}{\leftarrow \small{ subtract }} & & \\
& & & & -26x & +39 & & & & \\
& & & & -26x & +39 & & \color{red}{\leftarrow \small{ -13(2x-3) } } & & \\
& & & & -- & -- & & \color{red}{\leftarrow \small{ subtract } } & & \\
& & & & & & & & & \\
& & & & & 0 & & \color{red}{\leftarrow \small{ Remainder } } & &
\end{array} $
$(\displaystyle \frac{f}{g})(x)= x^2-3x-13$
