## Elementary Algebra

We know that if $x^{2}=a$, then $x=\pm \sqrt{a}$. Thus, we obtain: Step 1: $(x-3)^{2}=-18$ Step 2: $x-3=\pm \sqrt {-18}$ Step 3: $x-3=\pm \sqrt {-1\times9\times2}$ Step 4: $x-3=\pm (\sqrt {-1}\times\sqrt {9}\times\sqrt 2)$ Step 5: $x-3=\pm (i\times3\times\sqrt {2})$ [as $i=\sqrt {-1}$] Step 6: $x-3=\pm (3i\sqrt {2})$ Step 7: $x=3\pm (3i\sqrt {2})$ Step 8: $x=3+3i\sqrt {2}$ or $x=3-3i\sqrt {2}$ Therefore, the solution set is {$3-3i\sqrt {2},3+3i\sqrt {2}$}. Therefore, the solution set consists of two complex numbers, not two real numbers. This means that the statement in the question is false.