Решите систему уравнений: (x-2)(y+1)=36; x-2y=6.

\[(2y + 4)(y + 1) = 36\]

\[2y^{2} + 4y + 2y + 4 - 36 = 0\]

\[2y^{2} + 6y - 32 = 0\ \ \ \ \ |\ :2\]

\[y^{2} + 3y - 16 = 0\]

\[D = 9 + 64 = 73\]

\[y_{1,2} = \frac{- 3 \pm \sqrt{73}}{2};\]

\[x_{1,2} = 2 \cdot \frac{- 3 \pm \sqrt{73}}{2} + 6 =\]

\[= - 3 \pm \sqrt{73} + 6 = 3 \pm \sqrt{73}.\]

\[Ответ:\left( 3 \pm \sqrt{73};\frac{- 3 \pm \sqrt{73}}{2} \right).\]