Решите уравнение: (x^2+2x)^2+13*(x^2+2x)+12=0.

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

\[t^{2} + 13t + 12 = 0\]

\[D = 13^{2} - 4 \cdot 1 \cdot 12 = 169 =\]

\[= 169 - 48 = 121\]

\[t_{1} = \frac{- 3 + \sqrt{121}}{2} = \frac{- 13 + 11}{2} =\]

\[= \frac{- 2}{2} = - 1\]

\[t_{2} - \frac{- 13 - \sqrt{121}}{2} = \frac{- 13 - 11}{2} =\]

\[= \frac{- 24}{2} = - 12\]

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

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

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

\[x + 1 = 0\ \ \]

\[x = - 1.\]

\[2)\ x^{2} + 2x = - 12\ \ \ \]

\[x^{2} + 2x + 12 = 0\]

\[D = 2^{2} - 4 \cdot 1 \cdot 12 = 4 - 48 =\]

\[= - 44 < 0\]

\[нет\ решения.\]

\[Ответ:\ - 1.\]