jmc

We substitute $f(2)=5(2{)}^2-\frac{1}{2}+3=\frac{45}{2}$ and $g(2)=(2{)}^2-k=4-k$. So $f(2)-g(2)=2$ gives us $\frac{45}{2}-4+k=2$. Solving for $k$, we find $k=\frac{4}{2}-\frac{45}{2}+\frac{8}{2}$ so $\boxed{k=\frac{-33}{2}}$.