Answer
$\{-5\}$
Work Step by Step
Rewrite the equation so the root is isolated on one side (with a + sign)
Add $8$ to both sides.
$\sqrt{2x+19}=x+8$
For the root to be real, x must be such that $ x\geq-19/2.$
The RHS is nonnegative, so the LHS = $ x+8 \;$ is also nonnegative.
All solutions must satisfy
$ x\geq-8\qquad (*)$
Square both sides.
$2x+19=(x+8)^{2} $
$2x+19=x^{2}+16x+64$
$0=x^{2}+14x+45$
We factor this trinomial by finding two factors of $+45$ that add to $+14$; they are $+5$ and $+9$.
$0=(x+5)(x+9)$
$ x=-9 \;$ does not satisfy (*). We discard it.
$ x=-5 \;$ is a valid solution (satisfies (*) )
Check:
$\sqrt{2(-5)+19}-8=-5$
$\sqrt{9}-8=-5$
$3-8=-5$
The solution set is $\{-5\}$.
