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By Vieta, we know that $r+s+t=-\frac{-3}{2}=\frac{3}{2}$. Then, $$(r+s+t{)}^2={\left(\frac{3}{2}\right)}^2.$$Expanding gives us $$r^2+s^2+t^2+2rs+2st+2tr=\frac{9}{4}.$$We can rearrange to get $$r^2+s^2+t^2=\frac{9}{4}-2(rs+st+tr).$$We note that $rs+st+tr$ is the symmetric sum of roots taken two at a time. By Vieta, we know that $$rs+st+tr=\frac{4}{2}=2.$$Therefore, $$r^2+s^2+t^2=\frac{9}{4}-2(2)=\boxed{-\frac{7}{4}}.$$