Answer
{$\pm3\sqrt 3$}
Work Step by Step
$27-x^{2}=0$ can be rearranged as $x^{2}=27$
According to the square root property,
if $x^{2}=27$
then $x=\sqrt (27)=\sqrt (9\times3)=\sqrt 9\sqrt 3=3\sqrt 3$
or $x=-\sqrt (27)=-\sqrt (9\times3)=-\sqrt 9\sqrt 3=-3\sqrt 3$
Therefore, the solution set is {$-3\sqrt 3,+3\sqrt 3$} or {$\pm3\sqrt 3$}
