Решите неравенство: 1/3 x^2+1/3 x-4<0.

\[\frac{1}{3}x^{2} + \frac{1}{3}x - 4 < 0\ \ \ \ | \cdot 3\]

\[x^{2} + x - 12 < 0\]

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

\[= 1 + 48 = 49\]

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

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

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

\[= - \frac{8}{2} = - 4\]

\[Ответ:( - 4;3).\]