B) (t - 1)(t + 1) = 2 5t - 10\frac{1}{2}; г) -2(2 + 7) = (z - 2)(z + 2).

Решим уравнения:

1. в) $$(t - 1)(t + 1) = 2(5t - 10\frac{1}{2})$$
   $$t^2 - 1 = 10t - 21$$
   $$t^2 - 10t + 20 = 0$$
   $$D = (-10)^2 - 4 \cdot 1 \cdot 20 = 100 - 80 = 20$$
   $$t_1 = \frac{10 + \sqrt{20}}{2 \cdot 1} = \frac{10 + 2\sqrt{5}}{2} = 5 + \sqrt{5}$$
   $$t_2 = \frac{10 - \sqrt{20}}{2 \cdot 1} = \frac{10 - 2\sqrt{5}}{2} = 5 - \sqrt{5}$$
   Ответ: $$t_1 = 5 + \sqrt{5}$$, $$t_2 = 5 - \sqrt{5}$$
2. 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: $$z_1 = \frac{1}{2}$$, $$z_2 = -4$$