Решите систему уравнений: y=x^2+6x+7; y-2x=4.

\[\left\{ \begin{matrix} y = x^{2} + 6x + 7 \\ y - 2x = 4\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} + 6x + 7 - 2x = 4\]

\[x^{2} + 4x + 7 - 4 = 0\]

\[x^{2} + 4x + 3 = 0\]

\[D = 4^{2} - 4 \cdot 1 \cdot 3 = 16 - 12 = 4\]

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

\[= - 1\]

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

\[= - 3\]

\[x_{1} = - 1 \Longrightarrow \ \ \ \ \ \ y_{1} =\]

\[= ( - 1)^{2} + 6 \cdot ( - 1) + 7 =\]

\[= 1 - 6 + 7 = 2.\]

\[x_{2} = - 3 \Longrightarrow \ \ \ \ y_{2} =\]

\[= ( - 3)^{2} + 6 \cdot ( - 3) + 7 =\]

\[= 9 - 18 + 7 = - 2.\]

\[Ответ:( - 1;2),\ \ \ ( - 3;\ - 2).\]