jmc

49 is the number whose square root is 7, so we must have $$(x+3{)}^2=49.$$ Therefore, we must have $x+3=7$ or $x+3=-7$. The first equation gives us $x=4$ the second gives us $x=-10$. Both are solutions, so the sum of all possible values of $x$ is $4+(-10)=\boxed{-6}$.