Решите уравнение: (x^2-9)(x^2+2x-3)=0.

\[\left( x^{2} - 9 \right)\left( x^{2} + 2x - 3 \right) = 0\]

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

\[x - 3 = 0\]

\[x + 3 = 0\]

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

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

\[= 16\]

\[x = 3\]

\[x = - 3\]

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

\[= 1\]

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

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

\[Ответ:3;\ - 3;1.\]