jmc

Let $r$ be the common root, so $$\begin{array}{rl}{r}^2-7r+b& =0,\\ r^2+2r-2b& =0.\end{array}$$Then $2(r^2-7r+b)+(r^2+2r-2b)=0,$ which simplifies to $3r^2-12r=3r(r-4)=0.$ Hence, the possible values of $r$ are $\boxed{0,4}.$ (These can be realized when $b=0$ and $b=12,$ respectively.)