jmc

Let $r$ and $s$ be the roots of $2x^2-3x-11=0,$ so by Vieta's formulas, $r+s=\frac{3}{2}$ and $rs=-\frac{11}{2}.$

Then $$\begin{array}{rl}\frac{4r^3-4s^3}{r-s}& =\frac{4(r-s)(r^2+rs+s^2)}{r-s}\\ & =4(r^2+rs+s^2)\\ & =4[(r+s{)}^2-rs]\\ & =4\left[{\left(\frac{3}{2}\right)}^2+\frac{11}{2}\right]\\ & =\boxed{31}.\end{array}$$