\[3 \cdot (6 + y)^{2} + 3y \cdot y =\]
\[= 10y(6 + y)\]
\[3 \cdot \left( 36 + 12y + y^{2} \right) + 3y^{2} =\]
\[= 60y + 10y^{2}\]
\[108 + 36y + 3y^{2} + 3y^{2} =\]
\[= 60y + 10y^{2}\]
\[4y^{2} + 24y - 108 = 0\ \ \ \ \ |\ :4\]
\[y^{2} + 6y - 27 = 0\]
\[D_{1} = 9 + 27 = 36\]
\[y_{1} = - 3 + 6 = 3;\ \ \]
\[y_{2} = - 3 - 6 = - 9.\]
\[\left\{ \begin{matrix} y = 3 \\ x = 9 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \left\{ \begin{matrix} y = - 9 \\ x = - 3 \\ \end{matrix} \right.\ \]
\(Ответ:(9;3);( - 3; - 9).\)