Решите неравенство: 3x^2+5x-8>=0

\[3x^{2} + 5x - 8 \geq 0\]

\[D = 25 + 96 = 121\]

\[x_{1} = \frac{- 5 + 11}{6} = 1;\]

\[x_{2} = \frac{- 5 - 11}{6} = - \frac{16}{6} = - 2\frac{2}{3};\]

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

\[x \leq - 2\frac{2}{3};x \geq 1.\]

\[Ответ:x \leq - 2\frac{2}{3};x \geq 1.\]