520. Найдите корни уравнения: a) (x+3)(x-4) = -12; 6) 1⅔t + (2t + 1)(⅓t - 1) = 0; 3

520. Найдите корни уравнения:

1. a) $$(x+3)(x-4) = -12$$
   $$x^2 - 4x + 3x - 12 = -12$$
   $$x^2 - x - 12 + 12 = 0$$
   $$x^2 - x = 0$$
   $$x(x - 1) = 0$$
   $$x_1 = 0$$
   $$x - 1 = 0$$
   $$x_2 = 1$$
   
2. б) $$1\frac{2}{3}t + (2t + 1)(\frac{1}{3}t - 1) = 0$$
   $$\frac{5}{3}t + \frac{2}{3}t^2 - 2t + \frac{1}{3}t - 1 = 0$$
   $$\frac{2}{3}t^2 + \frac{5}{3}t - 2t + \frac{1}{3}t - 1 = 0$$
   $$\frac{2}{3}t^2 + \frac{6}{3}t - 2t - 1 = 0$$
   $$\frac{2}{3}t^2 + 2t - 2t - 1 = 0$$
   $$\frac{2}{3}t^2 - 1 = 0$$
   $$\frac{2}{3}t^2 = 1$$
   $$t^2 = 1 \cdot \frac{3}{2}$$
   $$t^2 = \frac{3}{2}$$
   $$t_1 = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$$
   $$t_2 = -\sqrt{\frac{3}{2}} = -\frac{\sqrt{6}}{2}$$
   

Ответ: а) $$x_1 = 0$$, $$x_2 = 1$$; б) $$t_1 = \frac{\sqrt{6}}{2}$$, $$t_2 = -\frac{\sqrt{6}}{2}$$