Решите уравнение: |x^2+2x-1|=2.

\[\left| x^{2} + 2x - 1 \right| = 2\]

\[x² + 2x - 1 = 2\]

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

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

\[= 16\]

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

\[= 1\]

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

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

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

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

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

\[x + 1 = 0\]

\[x = - 1.\]

\[Ответ:1;\ - 3;\ - 1.\]