Answer
{$\frac{2- 2\sqrt {7}}{3},\frac{2+2\sqrt {7}}{3}$}
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: $(3n-2)^{2}=28$
Step 2: $3n-2=\pm \sqrt {28}$
Step 3: $3n-2=\pm \sqrt {4\times7}$
Step 4: $3n-2=\pm \sqrt {2^{2}\times7}$
Step 5: $3n-2=\pm 2\sqrt {7}$
Step 6: $3n-2=+ 2\sqrt {7}$ or $3n-2=-2 \sqrt {7}$
Step 7: $3n=2+ 2\sqrt {7}$ or $3n=2-2 \sqrt {7}$
Step 8: $n=\frac{2+ 2\sqrt {7}}{3}$ or $n=\frac{2- 2\sqrt {7}}{3}$
The solution set is {$\frac{2- 2\sqrt {7}}{3},\frac{2+2\sqrt {7}}{3}$}.
