Answer
$\{6\}$
Work Step by Step
For the root to be real, x must be $\geq-10.$
The LHS is nonnegative, so the RHS = $ x-2$ is also nonnegative.
All solutions must satisfy
$ x\geq 2\qquad (*)$
$\sqrt{x+10}=x-2 \quad $... Square both sides
$ x+10=(x-2)^{2}$
$ x+10=x^{2}-4x+4$
$0=x^{2}-5x-6$
We factor this trinomial by finding two factors of $-6$ that add to $-5$; they are $-6$ and $+1$.
$(x-6)(x+1)=0$
$ x=6,\quad x=-1$
$ x=-1 \;$ is extraneous (does not satisfy (*) ), so we discard it.
$ x=6$ is a valid solution
Check:
$\sqrt{6+10}=6-2$
$ \sqrt{16}=4$
The solution set is $\{6\}$.
