Решите уравнение: (3x^2-2x-5)(x+2)=0.

\[\left( 3x^{2} - 2x - 5 \right)(x + 2) = 0\]

\[x + 2 = 0;\ \ \ x = - 2.\]

\[3x^{2} - 2x - 5 = 0\]

\[D = ( - 2)^{2} - 4 \cdot 3 \cdot ( - 5) =\]

\[= 4 + 60 = 64;\ \ \ \sqrt{D} = 8.\]

\[x_{1} = \frac{2 + 8}{2 \cdot 3} = \frac{10}{6} = \frac{5}{3} = 1\frac{2}{3};\ \ \ \ \ \ \ \]

\[\ x_{2} = \frac{2 - 8}{2 \cdot 3} = \frac{- 6}{6} = - 1\]

\[Ответ:\ - 2;1\frac{2}{3};\ - 1.\]