jmc

We begin by looking at each of the two possible cases; either $x<-5$ and $f(x)=x^2+9=10$, or $x\ge -5$ and $f(x)=3x-8=10$.

Tackling the first case, we find that the only possible values of $x$ that could satisfy $x^2+9=10\Rightarrow x^2=1$ are 1 and -1, neither of which are less than -5, thus yielding no possible solutions.

In the second case, the only possible value of $x$ that satisfies $3x-8=10$ is 6. Since this value is greater than or equal to -5, it satisfies both conditions. Thus, the only possible value of $x$ for which $f(x)=10$ is $6$, which means the sum of all possible values is also $\boxed{6}$.