Решите уравнение (x^2+3x+1)(x^2+3x-9)=171.

\[\left( x^{2} + 3x + 1 \right)\left( x^{2} + 3x - 9 \right) = 171\ \]

\[Пусть\ y = x^{2} + 3x + 1:\]

\[y(y - 10) = 171\]

\[y^{2} - 10y - 171 = 0\]

\[D = 25 + 171 = 196\]

\[y_{1} = 5 + 14 = 19;\ \ \ y_{2} = 5 - 14 = - 9.\]

\[Подставим:\]

\[1)\ x^{2} + 3x + 1 = 19\]

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

\[x_{1} + x_{2} = - 3;\ \ \ x_{1} \cdot x_{2} = - 18\]

\[x_{1} = - 6;\ \ \ x_{2} = 3.\]

\[2)\ x^{2} + 3x + 1 = - 9\]

\[x^{2} + 3x + 10 = 0\]

\[D = 9 - 40 = - 31 < 0\]

\[нет\ корней.\]

\[Ответ:x = - 6;\ \ x = 3.\]