jmc

If $x\ne -7,$ then we can cancel the factors of $x+7$ to get $$h(x)=2(x-3).$$If $x$ were allowed to be any real number, then $2(x-3)$ could also be any real number. However, the function is not defined for $x=-7,$ so the function cannot take on the value $2(-7-3)=-20.$

Therefore, the range of the function is $\boxed{(-\infty ,-20)\cup (-20,\infty )}.$