520. Найдите корни уравнения: a) (x + 3)(x – 4) = -12; б) 1 ⅔ t + (2t + 1)(⅓ t – 1) = 0;

Решим уравнения:

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

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

$$x^2 - x = 0$$

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

$$x_1 = 0$$

$$x - 1 = 0$$

$$x_2 = 1$$

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

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

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

$$2t^2 - 6t - 3 = 0$$

$$D = (-6)^2 - 4 \cdot 2 \cdot (-3) = 36 + 24 = 60$$

$$t_1 = \frac{6 + \sqrt{60}}{4} = \frac{6 + 2\sqrt{15}}{4} = \frac{3 + \sqrt{15}}{2}$$

$$t_2 = \frac{6 - \sqrt{60}}{4} = \frac{6 - 2\sqrt{15}}{4} = \frac{3 - \sqrt{15}}{2}$$

Ответ: a) $$x_1 = 0$$, $$x_2 = 1$$; б) $$t_1 = \frac{3 + \sqrt{15}}{2}$$, $$t_2 = \frac{3 - \sqrt{15}}{2}$$