Given \[ F(s)=\frac{2s^{2}-4}{(s-2)(s+1)(s-3)}. \] Decompose using partial fractions: \[ F(s)=\frac{A}{s-2}+\frac{B}{s+1}+\frac{C}{s-3}. \]
\[\begin{aligned} A &= \left.\frac{2s^2-4}{(s+1)(s-3)}\right|_{s=2} = \frac{4}{-3} = -\frac{4}{3},\\[4pt] B &= \left.\frac{2s^2-4}{(s-2)(s-3)}\right|_{s=-1} = \frac{-2}{12} = -\frac{1}{6},\\[4pt] C &= \left.\frac{2s^2-4}{(s-2)(s+1)}\right|_{s=3} = \frac{14}{4} = \frac{7}{2}. \end{aligned}\]
\[ F(s)= -\frac{\,4\,}{3}\,\frac{1}{s-2} -\frac{1}{6}\,\frac{1}{s+1} +\frac{7}{2}\,\frac{1}{s-3}. \]
Using the pair $\mathcal{L}^{-1}\{\tfrac{1}{s-a}\}=e^{at}$ and linearity, \[ f(t)= -\frac{4}{3}e^{2t} - \frac{1}{6}e^{-t} + \frac{7}{2}e^{3t}. \]