540. Решите уравнение: a) 5x2 = 9x + 2; 6)-t² 5t 14; в) 6x + 9 = x²; г) 2 - 5 = 2² - 25;

Решим уравнения:

1. a) $$5x^2 = 9x + 2$$
   $$5x^2 - 9x - 2 = 0$$
   $$D = (-9)^2 - 4 \cdot 5 \cdot (-2) = 81 + 40 = 121$$
   $$x_1 = \frac{9 + \sqrt{121}}{2 \cdot 5} = \frac{9 + 11}{10} = \frac{20}{10} = 2$$
   $$x_2 = \frac{9 - \sqrt{121}}{2 \cdot 5} = \frac{9 - 11}{10} = \frac{-2}{10} = -0.2$$
   Ответ: $$x_1 = 2$$, $$x_2 = -0.2$$
2. б) $$-t^2 = 5t - 14$$
   $$t^2 + 5t - 14 = 0$$
   $$D = 5^2 - 4 \cdot 1 \cdot (-14) = 25 + 56 = 81$$
   $$t_1 = \frac{-5 + \sqrt{81}}{2 \cdot 1} = \frac{-5 + 9}{2} = \frac{4}{2} = 2$$
   $$t_2 = \frac{-5 - \sqrt{81}}{2 \cdot 1} = \frac{-5 - 9}{2} = \frac{-14}{2} = -7$$
   Ответ: $$t_1 = 2$$, $$t_2 = -7$$
3. в) $$6x + 9 = x^2$$
   $$x^2 - 6x - 9 = 0$$
   $$D = (-6)^2 - 4 \cdot 1 \cdot (-9) = 36 + 36 = 72$$
   $$x_1 = \frac{6 + \sqrt{72}}{2 \cdot 1} = \frac{6 + 6\sqrt{2}}{2} = 3 + 3\sqrt{2}$$
   $$x_2 = \frac{6 - \sqrt{72}}{2 \cdot 1} = \frac{6 - 6\sqrt{2}}{2} = 3 - 3\sqrt{2}$$
   Ответ: $$x_1 = 3 + 3\sqrt{2}$$, $$x_2 = 3 - 3\sqrt{2}$$
4. г) $$z - 5 = z^2 - 25$$
   $$z^2 - z - 20 = 0$$
   $$D = (-1)^2 - 4 \cdot 1 \cdot (-20) = 1 + 80 = 81$$
   $$z_1 = \frac{1 + \sqrt{81}}{2 \cdot 1} = \frac{1 + 9}{2} = \frac{10}{2} = 5$$
   $$z_2 = \frac{1 - \sqrt{81}}{2 \cdot 1} = \frac{1 - 9}{2} = \frac{-8}{2} = -4$$
   Ответ: $$z_1 = 5$$, $$z_2 = -4$$