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

a) $$(2x - 3)(5x + 1) = 2x + \frac{2}{5}$$

$$10x^2 + 2x - 15x - 3 = 2x + \frac{2}{5}$$

$$10x^2 - 13x - 3 - 2x - \frac{2}{5} = 0$$

$$10x^2 - 15x - \frac{17}{5} = 0$$

$$50x^2 - 75x - 17 = 0$$

$$D = (-75)^2 - 4 \cdot 50 \cdot (-17) = 5625 + 3400 = 9025$$

$$x_1 = \frac{75 + \sqrt{9025}}{2 \cdot 50} = \frac{75 + 95}{100} = \frac{170}{100} = 1.7$$

$$x_2 = \frac{75 - \sqrt{9025}}{2 \cdot 50} = \frac{75 - 95}{100} = \frac{-20}{100} = -0.2$$

Ответ: $$x_1 = 1.7$$, $$x_2 = -0.2$$

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: $$y_1 = \frac{7 + \sqrt{13}}{6}$$, $$y_2 = \frac{7 - \sqrt{13}}{6}$$