Решите неравенство: -3x^2+4x+4<=0.

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

\[3x^{2} - 4x - 4 \geq 0\]

\[D_{1} = 4 + 12 = 16\]

\[x_{1} = \frac{2 + 4}{3} = 2;\ \]

\[x_{2} = \frac{2 - 4}{3} = - \frac{2}{3}.\]

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

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

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