jmc

Using the quadratic formula, we see that the solutions of the quadratic equation $x^2+bx+(b+3)=0$ are given by $\frac{-b\pm \sqrt{b^2-4(b+3)}}{2}$. So we may set $\frac{-b+\sqrt{b^2-4(b+3)}}{2}$ equal to $\frac{-b+\sqrt{5}}{2}$ which implies $b^2-4b-12=5⟹b^2-4b-17=0$. (Note that setting $\frac{-b+\sqrt{b^2-4(b+3)}}{2}$ equal to $\frac{-b-\sqrt{5}}{2}$ gives no solution). We must use the quadratic formula again. We get $$b=\frac{4\pm \sqrt{4^2-4(-17)}}{2}=\frac{4\pm \sqrt{84}}{2}=2\pm \sqrt{21}.$$Take the positive root and sum: $m+n=2+21=\boxed{23}$.