Question

36=(s+2)^2

Answer 1

s=-8 or s=4

Answer 2

\[(s+2)^2=36\]take square root\[s+2 = \pm \sqrt{36}\]subtract 2; evaluate root\[s=-2 \pm 6\]write solutions\[s=-2-6=-8 \cup s=-2+6=4\]solution set is {-8, 4}

Answer 3

thank you