solve the following quadratic equation (x+3)2+25=0
\(2\cdot(x+3)+25=0\) ⇒ \(2x+2\cdot3=0-25\)\(2x+6=-25\) ⇒ \(2x=-25-6\) ⇒ \(2x=-31\)\(x= \frac{-31}{2}\) ⇒ \(\boxed {x=-15,5}\)
