\[ s^2-3s+2=(s-1)(s-2),\qquad F(s)=\frac{3s-4}{(s-1)(s-2)}. \]
Assume \[ F(s)=\frac{A}{s-1}+\frac{B}{s-2} \] so that \[ 3s-4=A(s-2)+B(s-1). \] Setting $s=1$ gives $-1=-A\Rightarrow A=1$, and $s=2$ gives $2=B$. Hence \[ F(s)=\frac{1}{s-1}+\frac{2}{s-2}. \]
Using $\mathcal{L}^{-1}\{\tfrac{1}{s-a}\}=e^{at}$ and linearity, \[ \begin{aligned} f(t) &= \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\!\left\{\frac{1}{s-1}\right\} + 2\,\mathcal{L}^{-1}\!\left\{\frac{1}{s-2}\right\} \\ &= e^{t}+2e^{2t}. \end{aligned} \]
\[ \boxed{\,f(t)=e^{t}+2e^{2t}\,} \]