smc

We factor $x^2-3x+2$ into $(x-1)(x-2)$. Thus, either $1$ or $2$ is a root of $x^2-5x+k$. If $1$ is a root, then $1^2-5\cdot 1+k=0$, so $k=4$. If $2$ is a root, then $2^2-5\cdot 2+k=0$, so $k=6$. The sum of all possible values of $k$ is $\boxed{10}$.