542. Найдите корни уравнения: a) (2x-3)(5x + 1) = б) (Зу - 1)(у + 3) = y(1 + 6y);

Решим уравнения:

1. a) $$(2x-3)(5x + 1) = 2x + \frac{2}{5}$$
   $$10x^2 + 2x - 15x - 3 = 2x + \frac{2}{5}$$
   $$10x^2 - 15x - 3 - \frac{2}{5} = 0$$
   $$10x^2 - 15x - \frac{17}{5} = 0$$
   $$50x^2 - 75x - 17 = 0$$
   $$D = (-75)^2 - 4 \cdot 50 \cdot (-17) = 5625 + 3400 = 9025$$
   $$x_1 = \frac{75 + \sqrt{9025}}{2 \cdot 50} = \frac{75 + 95}{100} = \frac{170}{100} = \frac{17}{10} = 1.7$$
   $$x_2 = \frac{75 - \sqrt{9025}}{2 \cdot 50} = \frac{75 - 95}{100} = \frac{-20}{100} = -0.2$$
   Ответ: $$x_1 = 1.7$$, $$x_2 = -0.2$$
2. б) $$(3y - 1)(y + 3) = y(1 + 6y)$$
   $$3y^2 + 9y - y - 3 = y + 6y^2$$
   $$3y^2 + 8y - 3 - y - 6y^2 = 0$$
   $$-3y^2 + 7y - 3 = 0$$
   $$3y^2 - 7y + 3 = 0$$
   $$D = (-7)^2 - 4 \cdot 3 \cdot 3 = 49 - 36 = 13$$
   $$y_1 = \frac{7 + \sqrt{13}}{2 \cdot 3} = \frac{7 + \sqrt{13}}{6}$$
   $$y_2 = \frac{7 - \sqrt{13}}{2 \cdot 3} = \frac{7 - \sqrt{13}}{6}$$
   Ответ: $$y_1 = \frac{7 + \sqrt{13}}{6}$$, $$y_2 = \frac{7 - \sqrt{13}}{6}$$