Решите систему уравнений x-y=2; y^2-3x=12.

\[\left\{ \begin{matrix} x - y = 2\ \ \ \ \ \ \ \\ y^{2} - 3x = 12 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 3 \cdot (2 + y) = 12 \\ \end{matrix} \right.\ \]

\[y^{2} - 6 - 3y - 12 = 0\]

\[y^{2} - 3y - 18 = 0\]

\[y_{1} + y_{2} = 3;\ \ y_{1} \cdot y_{2} = - 18\]

\[y_{1} = 6;\ \ \ \ \ \ \ y_{2} = - 3.\]

\[\left\{ \begin{matrix} y = 6\ \ \ \ \ \ \ \ \\ x = 2 + 6 \\ \end{matrix} \right.\ \Leftrightarrow \left\{ \begin{matrix} y = - 3\ \ \ \ \\ x = 2 - 3 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 6 \\ x = 8 \\ \end{matrix} \right.\ \Leftrightarrow \left\{ \begin{matrix} y = - 3 \\ x = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:(8;6)\ или\ ( - 1; - 3).\]