Решите уравнение: (x^2+3x)^3-14x^2-42x+40=0.

\[\left( x^{2} + 3x \right)^{2} - 14x^{2} - 42x + 40 = 0\]

\[\left( x^{2} + 3x \right)^{2} - 14 \cdot \left( x^{2} + 3x \right) + 40 = 0\]

\(t = x^{2} + 3x\)

\[t^{2} - 14t + 40 = 0\]

\[D = ( - 14)^{2} - 4 \cdot 1 \cdot 40 =\]

\[= 196 - 160 = 36;\ \ \ \ \sqrt{D} = 6.\]

\[t_{1} = \frac{14 + 6}{2} = \frac{20}{2} = 10;\ \ \ \]

\[\text{\ \ \ \ }t_{2} = \frac{14 - 6}{2} = \frac{8}{2} = 4\]

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

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

\[D = 3^{2} - 4 \cdot 1 \cdot ( - 10) =\]

\[= 9 + 40 = 49;\ \ \ \sqrt{D} = 7.\]

\[x_{1} = \frac{- 3 + 7}{2} = \frac{4}{2} = 2;\ \ \ \ \ \ \ \]

\[\text{\ \ \ \ }x_{2} = \frac{- 3 - 7}{2} = \frac{- 10}{2} = - 5\]

\[x^{2} + 3x = 4\]

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

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

\[= 9 + 16 = 25;\ \ \ \ \sqrt{D} = 5.\]

\[x_{1} = \frac{- 3 + 5}{2} = \frac{2}{2} = 1;\ \ \ \ \]

\[\text{\ \ \ \ }x_{2} = \frac{- 3 - 5}{2} = \frac{- 8}{2} = - 4\]

\[Ответ:2;\ - 5;\ 1;\ - 4.\]