Найдите корни уравнения: (x+3)^2=2x+6.

\[(x + 3)^{2} = 2x + 6\]

\[x^{2} + 6x + 9 = 2x + 6\]

\[x^{2} + 6x + 9 - 2x - 6 = 0\]

\[x² + 4x + 3 = 0\]

\[D = b^{2} - 4ac = 16 - 4 \cdot 1 \cdot 3 =\]

\[= 16 - 12 = 4\]

\[x_{1} = \frac{- 4 + 2}{2} = - \frac{2}{2} = - 1\]

\[x_{2} = \frac{- 4 - 2}{2} = - \frac{6}{2} = - 3\]

\[Ответ:x_{1} = - 1;\ \ x_{2} = - 3.\]