Решите неравенство: 3x^2+11x-4<0.

Ответ:

\[3x^{2} + 11x - 4 < 0\]

\[3x^{2} + 11x - 4 = 0\]

\[D = 121 + 48 = 169\]

\[x_{1} = \frac{- 11 + 13}{6} = \frac{2}{6} = \frac{1}{3};\ \ \ \]

\[x_{2} = \frac{- 11 - 13}{6} = - 4\]

\[(x + 4)\left( x - \frac{1}{3} \right) < 0\]

\[Ответ:\ x \in \left( - 4;\frac{1}{3} \right).\]