Решите квадратное уравнение: 1) $$7x^2 - 9x + 2 = 0$$; 2) $$7x^2 - 28 = 0$$; 3) $$5x^2 = 12x$$; 4) $$x^2 + 20x + 91 = 0$$.

1) $$7x^2 - 9x + 2 = 0$$ $$D = (-9)^2 - 4 * 7 * 2 = 81 - 56 = 25$$ $$x_1 = \frac{9 + \sqrt{25}}{2 * 7} = \frac{9 + 5}{14} = \frac{14}{14} = 1$$ $$x_2 = \frac{9 - \sqrt{25}}{2 * 7} = \frac{9 - 5}{14} = \frac{4}{14} = \frac{2}{7}$$ Ответ: $$x_1 = 1$$, $$x_2 = \frac{2}{7}$$ 2) $$7x^2 - 28 = 0$$ $$7x^2 = 28$$ $$x^2 = 4$$ $$x = \pm \sqrt{4}$$ $$x_1 = 2$$, $$x_2 = -2$$ Ответ: $$x_1 = 2$$, $$x_2 = -2$$ 3) $$5x^2 = 12x$$ $$5x^2 - 12x = 0$$ $$x(5x - 12) = 0$$ $$x_1 = 0$$ $$5x - 12 = 0$$ $$5x = 12$$ $$x_2 = \frac{12}{5} = 2.4$$ Ответ: $$x_1 = 0$$, $$x_2 = 2.4$$ 4) $$x^2 + 20x + 91 = 0$$ $$D = 20^2 - 4 * 1 * 91 = 400 - 364 = 36$$ $$x_1 = \frac{-20 + \sqrt{36}}{2 * 1} = \frac{-20 + 6}{2} = \frac{-14}{2} = -7$$ $$x_2 = \frac{-20 - \sqrt{36}}{2 * 1} = \frac{-20 - 6}{2} = \frac{-26}{2} = -13$$ Ответ: $$x_1 = -7$$, $$x_2 = -13$$