Решите уравнение: 2x^3-3x^2-3x+2=0.

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

\[2 \bullet \left( x^{3} + 1 \right) - 3x(x + 1) = 0\]

\[(x + 1)\left( 2x^{2} - 2x + 2 - 3x \right) = 0\]

\[(x + 1)\left( 2x^{2} - 5x + 2 \right) = 0\]

\[x + 1 = 0 \Longrightarrow x = - 1.\]

\[2x^{2} - 5x + 2 = 0\]

\[D = ( - 5)^{2} - 4 \cdot 2 \cdot 2 =\]

\[= 25 - 16 = 9\]

\[x_{1} = \frac{5 + \sqrt{9}}{2 \cdot 2} = \frac{5 + 3}{4} = \frac{8}{4} = 2\]

\[x_{2} = \frac{5 - \sqrt{9}}{2 \cdot 2} = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2}\]

\[Ответ:\ x = - 1;\ x = 2;\ \ x = 0,5.\]