542. Найдите корни уравнения: в) (t – 1)(t + 1) = 2( 5t-10/2 ) г) -z(z + 7) = (z – 2)(z+2).

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

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

$$t^2 - 10t + 9 = 0$$

$$D = (-10)^2 - 4 \cdot 1 \cdot 9 = 100 - 36 = 64$$

$$t_1 = \frac{10 + \sqrt{64}}{2 \cdot 1} = \frac{10 + 8}{2} = \frac{18}{2} = 9$$

$$t_2 = \frac{10 - \sqrt{64}}{2 \cdot 1} = \frac{10 - 8}{2} = \frac{2}{2} = 1$$

Ответ: $$t_1 = 9$$, $$t_2 = 1$$

г) $$-z(z + 7) = (z - 2)(z + 2)$$

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

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

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

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

$$z_1 = \frac{-7 + \sqrt{81}}{2 \cdot 2} = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2}$$

$$z_2 = \frac{-7 - \sqrt{81}}{2 \cdot 2} = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$$

Ответ: $$z_1 = \frac{1}{2}$$, $$z_2 = -4$$