Key concept: If $b^x = b^y$, then $x = y$ (as long as $b \neq 0$, $b \neq 1$, and $b \neq -1$)

The provisos here a very important.
For example, if $0^x = 0^y$, we can't then conclude that $x=y$

Now on to the solution.....
If $x^{(x^2 + 2)} = x^{(4x + 14)}$, then we can conclude that: $x^2 + 2 = 4x + 14$
Rewrite as follows: $x^2 - 4x - 12= 0$
Factor: $(x-6)(x+2)=0$
So, two solutions are $x=6$ and $x = -2$

At this point, we need to test the "proviso" values ($x=0$, $x=1$ and $x=-1$)

Test $x=0$ to get: $0^{0^2 + 2} = 0^{4(0) + 14}$. Works! So, another solution is $x=0$
Test $x=1$ to get: $1^{1^2 + 2} = 1^{4(1) + 14}$. Works! So, another solution is $x=1$
Test $x=-1$ to get: $(-1)^{(-1)^2 + 2} = (-1)^{4(-1) + 14}$. DOESN'T work, So, $x=-1$ is NOT a solution

So, the sum of the solutions $= 6 + (-2) + 0 + 1 = 5$

Answer: C