в) 3y² + 4y – 4 > 0;

в) $$3y^2 + 4y - 4 > 0$$

$$3y^2 + 4y - 4 = 0$$

$$D = 16 + 48 = 64$$

$$y_1 = \frac{-4 + 8}{6} = \frac{2}{3}$$

$$y_2 = \frac{-4 - 8}{6} = -2$$

$$3(y - \frac{2}{3})(y + 2) > 0$$

$$(y - \frac{2}{3})(y + 2) > 0$$

___-2____2/3____

+ - +

$$y \in (-\infty; -2) \cup (\frac{2}{3}; +\infty)$$

Ответ: $$y \in (-\infty; -2) \cup (\frac{2}{3}; +\infty)$$