Answer
{$-\sqrt {14},\sqrt {14}$}
Work Step by Step
Using Property 10.1, which states that for any non-negative real number $a$, $x^{2}=a$ can be written as $x=\pm\sqrt a$, we obtain:
Step 1: $n^{2}=14$
Step 2: $n=\pm \sqrt {14}$
The solution set is {$-\sqrt {14},\sqrt {14}$}.
