Answer
$\{10\}$
Work Step by Step
Rewrite the equation so the root is isolated on one side (with a + sign).
Add $\sqrt{2x+5}-5$ to both sides.
$ x-5=\sqrt{2x+5}$
For the root to be real, x must be such that $ x\geq-5/2.$
The RHS is nonnegative, so the LHS = $ x-5 \;$ is also nonnegative.
All solutions must satisfy
$ x\geq 5\qquad (*)$
Square both sides.
$(x-5)^{2}=2x+5$
$ x^{2}-10x+25=2x+5$
$ x^{2}-12x+20=0$
We factor this trinomial by finding two factors of $+20$ that add to $-12$; they are $-10$ and $-2$.
$(x-2)(x-10)=0$
$ x=2 \;$ does not satisfy (*). We discard it.
$ x=10 \;$ is a valid solution (satisfies (*) )
Check:
$10-\sqrt{2(10)+5}=5$
$10-\sqrt{25}=5$
$10-5=5$
The solution set is $\{10\}$.
