#### Answer

$\{-3\}$

#### Work Step by Step

Rewrite the equation so the root is isolated on one side (with a + sign)
Add $6$ to both sides.
$\sqrt{2x+15}=x+6$
For the root to be real, x must be such that $ x\geq-15/2.$
The RHS is nonnegative, so the LHS = $ x+6 \;$ is also nonnegative.
All solutions must satisfy
$ x\geq-6\qquad (*)$
Square both sides.
$2x+15=x^{2}+12x+36$
$0=x^{2}+10x+21$
We factor this trinomial by finding two factors of $+21$ that add to $+10$; they are $+3$ and $+7$.
$0=(x+3)(x+7)$
$ x=-7 \;$ does not satisfy (*). We discard it.
$ x=-3 \;$ is a valid solution (satisfies (*) )
Check:
$\sqrt{2(-3)+15}-6=-3$
$\sqrt{9}-6=-3$
$3-6=-3$
The solution set is $\{-3\}$.