542. Найдите корни уравнения: a) $$(2x-3)(5x + 1) = 2x+\frac{2}{5}$$; б) $$(3y-1)(y + 3) = y(1 + 6y)$$; в) $$(t - 1)(t + 1) = 2(5t - 10\frac{1}{2})$$; г) $$-z(z + 7) = (z - 2)(z + 2)$$.

Решим уравнения:

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

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

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

$$50x^2 - 65x - 15 = 10x + 2$$

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

$$D = (-75)^2 - 4 \cdot 50 \cdot (-17) = 5625 + 3400 = 9025 = 95^2$$

$$x_1 = \frac{75 + 95}{100} = \frac{170}{100} = 1.7$$

$$x_2 = \frac{75 - 95}{100} = \frac{-20}{100} = -0.2$$

Ответ: $$x_1 = 1.7, x_2 = -0.2$$

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б) $$(3y-1)(y + 3) = y(1 + 6y)$$

$$3y^2 + 9y - y - 3 = y + 6y^2$$

$$3y^2 + 8y - 3 = y + 6y^2$$

$$3y^2 - 7y + 3 = 0$$

$$D = (-7)^2 - 4 \cdot 3 \cdot 3 = 49 - 36 = 13$$

$$y_1 = \frac{7 + \sqrt{13}}{6}$$

$$y_2 = \frac{7 - \sqrt{13}}{6}$$

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

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в) $$(t - 1)(t + 1) = 2(5t - 10\frac{1}{2})$$

$$t^2 - 1 = 10t - 21$$

$$t^2 - 10t + 20 = 0$$

$$D = (-10)^2 - 4 \cdot 1 \cdot 20 = 100 - 80 = 20$$

$$t_1 = \frac{10 + \sqrt{20}}{2} = \frac{10 + 2\sqrt{5}}{2} = 5 + \sqrt{5}$$

$$t_2 = \frac{10 - \sqrt{20}}{2} = \frac{10 - 2\sqrt{5}}{2} = 5 - \sqrt{5}$$

Ответ: $$t_1 = 5 + \sqrt{5}, t_2 = 5 - \sqrt{5}$$

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г) $$-z(z + 7) = (z - 2)(z + 2)$$

$$-z^2 - 7z = z^2 - 4$$

$$2z^2 + 7z - 4 = 0$$

$$D = 7^2 - 4 \cdot 2 \cdot (-4) = 49 + 32 = 81$$

$$z_1 = \frac{-7 + \sqrt{81}}{4} = \frac{-7 + 9}{4} = \frac{2}{4} = 0.5$$

$$z_2 = \frac{-7 - \sqrt{81}}{4} = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$$

Ответ: $$z_1 = 0.5, z_2 = -4$$