Решите неравенство: 1/3 x^2+x-10/3<0.

\[\frac{1}{3}x^{2} + x - \frac{10}{3} < 0\ \ \ \ | \cdot 3\]

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

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

\[= 9 + 40 = 49\]

\[x_{1} = \frac{- 3 + \sqrt{49}}{2 \cdot 1} = \frac{- 3 + 7}{2} =\]

\[= \frac{4}{2} = 2;\ \ \ \ \ \]

\[x_{2} = \frac{- 3 - \sqrt{49}}{2 \cdot 1} = \frac{- 3 - 7}{2} =\]

\[= - \frac{10}{2} = - 5\]

\[Ответ:( - 5;2).\]