Answer
{$-1-2\sqrt 3,-1+2\sqrt 3$}
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: $(n+1)^{2}=12$
Step 2: $n+1=\pm \sqrt {12}$
Step 3: $n+1=\pm \sqrt {4\times3}$
Step 4: $n+1=\pm \sqrt {2^{2}\times3}$
Step 5: $n+1=\pm 2\sqrt {3}$
Step 6: $n+1=+ 2\sqrt {3}$ or $n+1=-2 \sqrt {3}$
Step 7: $n=-1+2\sqrt 3$ or $n=-1-2\sqrt 3$
The solution set is {$-1-2\sqrt 3,-1+2\sqrt 3$}.
