344. Используя формулу корней квадратного уравнения с чётным вторым коэффициентом, решите уравнение:
1) $$x^2 - 16x + 63 = 0$$
$$D_1 = (\frac{-16}{2})^2 - 63 = 64 - 63 = 1$$
$$x_1 = 8 + 1 = 9, x_2 = 8 - 1 = 7$$
2) $$y^2 + 14y + 48 = 0$$
$$D_1 = (\frac{14}{2})^2 - 48 = 49 - 48 = 1$$
$$y_1 = -7 + 1 = -6, y_2 = -7 - 1 = -8$$
3) $$z^2 + 24z + 150 = 0$$
$$D_1 = (\frac{24}{2})^2 - 150 = 144 - 150 = -6 < 0$$. Вещественных корней нет.
4) $$x^2 - 20x - 125 = 0$$
$$D_1 = (\frac{-20}{2})^2 + 125 = 100 + 125 = 225$$
$$x_1 = 10 + 15 = 25, x_2 = 10 - 15 = -5$$
5) $$5p^2 + 6p - 8 = 0$$
$$D_1 = (\frac{6}{2})^2 - 5 \cdot (-8) = 9 + 40 = 49$$
$$p_1 = \frac{-3 + 7}{5} = \frac{4}{5}, p_2 = \frac{-3 - 7}{5} = -2$$
6) $$7k^2 - 20k + 14 = 0$$
$$D_1 = (\frac{-20}{2})^2 - 7 \cdot 14 = 100 - 98 = 2$$
$$k_1 = \frac{10 + \sqrt{2}}{7}, k_2 = \frac{10 - \sqrt{2}}{7}$$
7) $$15t^2 - 22t - 37 = 0$$
$$D_1 = (\frac{-22}{2})^2 - 15 \cdot (-37) = 121 + 555 = 676$$
$$t_1 = \frac{11 + 26}{15} = \frac{37}{15}, t_2 = \frac{11 - 26}{15} = -1$$
8) $$35m^2 + 36m + 1 = 0$$
$$D_1 = (\frac{36}{2})^2 - 35 \cdot 1 = 324 - 35 = 289$$
$$m_1 = \frac{-18 + 17}{35} = -\frac{1}{35}, m_2 = \frac{-18 - 17}{35} = -1$$
Ответ: 1) $$x_1 = 9, x_2 = 7$$; 2) $$y_1 = -6, y_2 = -8$$; 3) Вещественных корней нет; 4) $$x_1 = 25, x_2 = -5$$; 5) $$p_1 = \frac{4}{5}, p_2 = -2$$; 6) $$k_1 = \frac{10 + \sqrt{2}}{7}, k_2 = \frac{10 - \sqrt{2}}{7}$$; 7) $$t_1 = \frac{37}{15}, t_2 = -1$$; 8) $$m_1 = -\frac{1}{35}, m_2 = -1$$