Answer
$\{6\}$
Work Step by Step
The LHS is nonnegative, so the RHS = $ x-3$ is also nonnegative.
All solutions must satisfy
$ x\geq 3\qquad (*)$
$\sqrt{x+3}=x-3 \quad $... Square both sides
$ x+3=x^{2}-6x+9$
$ x^{2}-7x+6=0$
We factor this trinomial by finding two factors of $+6$ that add to $-7$; 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:
$\sqrt{6+3}=6-3$
$ \sqrt{9}=3$
The solution set is $\{6\}$.
