520. Найдите корни уравнения: a) (x + 3)(x - 4) = −12; б) 1⅔t + (2t + 1)(⅓t - 1) = 0;

a) $$(x + 3)(x - 4) = -12$$

$$x^2 - 4x + 3x - 12 = -12$$

$$x^2 - x = 0$$

$$x(x - 1) = 0$$

$$x_1 = 0, x_2 = 1$$

Ответ: x₁ = 0, x₂ = 1

б) $$1\frac{2}{3}t + (2t + 1)(\frac{1}{3}t - 1) = 0$$

$$\frac{5}{3}t + \frac{2}{3}t^2 - 2t + \frac{1}{3}t - 1 = 0$$

$$\frac{2}{3}t^2 + (\frac{5}{3} - 2 + \frac{1}{3})t - 1 = 0$$

$$\frac{2}{3}t^2 + (\frac{6}{3} - 2)t - 1 = 0$$

$$\frac{2}{3}t^2 - 1 = 0$$

$$\frac{2}{3}t^2 = 1$$

$$t^2 = \frac{3}{2}$$

$$t_1 = \sqrt{\frac{3}{2}}, t_2 = -\sqrt{\frac{3}{2}}$$

Ответ: $$t_1 = \sqrt{\frac{3}{2}}, t_2 = -\sqrt{\frac{3}{2}}$$, или $$t_1 = \frac{\sqrt{6}}{2}, t_2 = -\frac{\sqrt{6}}{2}$$