Решите уравнение: (2x^2+4x-5)(x^2+2x-2)-5x^2-10x-26=0.

\[t = x^{2} + 2x - 2\]

\[(2t - 1)t - (5t + 36) = 0\]

\[2t^{2} - t - 5t - 36 = 0\]

\[2t^{2} - 6t - 36 = 0\ \ \ \ \ |\ :2\]

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

\[D = ( - 3)^{2} - 4 \cdot 1 \cdot ( - 18) =\]

\[= 9 + 72 = 81\]

\[t_{1} = \frac{3 + \sqrt{81}}{2} = \frac{3 + 9}{2} = \frac{12}{2} = 6\]

\[t_{2} = \frac{3 - \sqrt{81}}{2} = \frac{3 - 9}{2} = \frac{- 6}{2} =\]

\[= - 3\]

\[1)\ x^{2} + 2x - 2 = 6\]

\[x^{2} + 2x - 8 = 0\]

\[D =^{2} - 4 \cdot 1 \cdot ( - 8) = 4 + 32 =\]

\[= 36\]

\[x_{1} = \frac{- 2 + \sqrt{36}}{2} = \frac{- 2 + 6}{2} = \frac{4}{2} =\]

\[= 2\]

\[x_{2} = \frac{- 2 - \sqrt{36}}{2} = \frac{- 2 - 6}{2} =\]

\[= \frac{- 8}{2} = - 4\]

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

\[x^{2} + 2x + 1 = 0\]

\[(x + 1)^{2} = 0\]

\[x + 1 = 0\]

\[x = - 1\]

\[Ответ:2;\ - 4;\ - 1.\]