Answer
$\{-6\}$
Work Step by Step
For the root to be real, x must be such that $ x\geq-13/2.$
The LHS is nonnegative, so the RHS = $ x+7 \;$ is also nonnegative.
All solutions must satisfy
$ x\geq-13/2\qquad (*)$
$2x+13=(x+7)^{2}$
$2x+13=x^{2}+14x+49$
$ x^{2}+12x+36=0 \quad $... recognize a perfect square
$(x+6)^{2}=0$
$ x=-6 \;$ is a valid solution (satisfies (*) )
Check:
$\sqrt{2(-6)+13}=-6+7$
$\sqrt{-12+13}=1$
$\sqrt{1}=1$
The solution set is $\{-6\}$.
