Решите уравнение: (2x^2-4x-3)(x^2-2x+3)-5x^2+10x-3=0.

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

\[(2t - 9)t - (5t - 12) = 0\]

\[2t^{2} - 9t - 5t + 12 = 0\]

\[2t^{2} - 14t + 12 = 0\ \ \ \ \ \ \ \ \ \ |\ :2\]

\[t^{2} - 7t + 6 = 0\]

\[D = ( - 7)^{2} - 4 \cdot 1 \cdot 6 =\]

\[= 49 - 24 = 25\]

\[t_{1} = \frac{7 + \sqrt{25}}{2} = \frac{7 + 5}{2} = \frac{12}{2} = 6\]

\[t_{2} = \frac{7 - \sqrt{25}}{2} = \frac{7 - 5}{2} = \frac{2}{2} = 1\]

\[1)\ x^{2} - 2x + 3 = 6\]

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

\[= 4 + 12 = 16\]

\[x_{1} = \frac{2 + \sqrt{16}}{2} = \frac{2 + 4}{2} = \frac{6}{2} = 3\]

\[x_{2} = \frac{2 - \sqrt{16}}{2} = \frac{2 - 4}{2} = \frac{- 2}{2} =\]

\[= - 1\]

\[2)\ x^{2} - 2x + 3 = 1\]

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

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

\[= - 4 < 0 \Longrightarrow нет\ решения.\]

\[Ответ:3;\ - 1.\]