- a) $$2x^2 - 5x = 0$$
$$x(2x - 5) = 0$$
$$x_1 = 0$$
$$2x - 5 = 0$$
$$2x = 5$$
$$x_2 = \frac{5}{2} = 2,5$$
- б) $$3y + 18y^2 = 0$$
$$y(3 + 18y) = 0$$
$$y_1 = 0$$
$$3 + 18y = 0$$
$$18y = -3$$
$$y_2 = -\frac{3}{18} = -\frac{1}{6}$$
Ответ: a) $$x_1 = 0, x_2 = 2,5$$, б) $$y_1 = 0, y_2 = -\frac{1}{6}$$