Решите уравнение: (2x^2+4x+1)(x^2+2x+1)-5x^2-10x-13=0.

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

\[(2t - 1)t - (5t + 8) = 0\]

\[1)\ 2t^{2} - t - 5t - 8 = 0\]

\[2t^{2} - 6t - 8 = 0\ \ \ \ \ |\ :2\]

\[t^{2} - 3t - 4 = 0\]

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

\[= 9 + 16 = 25\]

\[t_{1} = \frac{3 + \sqrt{25}}{2} = \frac{3 + 5}{2} = \frac{8}{2} = 4\]

\[t_{2} = \frac{3 - \sqrt{25}}{2} = \frac{3 - 5}{2} = \frac{- 2}{2} =\]

\[= - 1\ (не\ подходит).\]

\[2)\ x^{2} + 2x + 1 = 4\]

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

\[D = 2^{2} - 4 \cdot 1 \cdot ( - 3) = 4 + 12 =\]

\[= 16\]

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

\[= 1\]

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

\[= \frac{- 6}{2} = - 3\]

\[Ответ:1;\ - 3.\]