540. Решите уравнение:

Для решения квадратных уравнений вида $ax^2 + bx + c = 0$ используем формулу дискриминанта $D = b^2 - 4ac$ и корни $x = \frac{-b \pm \sqrt{D}}{2a}$. а) $5x^2 - 9x - 2 = 0$ $D = (-9)^2 - 4 \cdot 5 \cdot (-2) = 81 + 40 = 121 = 11^2$ $x_1 = \frac{9+11}{10} = 2$, $x_2 = \frac{9-11}{10} = -0,2$ б) $-t^2 - 5t + 14 = 0 \Rightarrow t^2 + 5t - 14 = 0$ $D = 5^2 - 4 \cdot 1 \cdot (-14) = 25 + 56 = 81 = 9^2$ $t_1 = \frac{-5+9}{2} = 2$, $t_2 = \frac{-5-9}{2} = -7$ в) $x^2 - 6x - 9 = 0$ $D = (-6)^2 - 4 \cdot 1 \cdot (-9) = 36 + 36 = 72 \Rightarrow \sqrt{D} = 6\sqrt{2}$ $x = \frac{6 \pm 6\sqrt{2}}{2} = 3 \pm 3\sqrt{2}$ г) $z^2 - z - 20 = 0$ $D = (-1)^2 - 4 \cdot 1 \cdot (-20) = 1 + 80 = 81 = 9^2$ $z_1 = \frac{1+9}{2} = 5$, $z_2 = \frac{1-9}{2} = -4$ д) $y^2 - 52y + 576 = 0$ $D = (-52)^2 - 4 \cdot 1 \cdot 576 = 2704 - 2304 = 400 = 20^2$ $y_1 = \frac{52+20}{2} = 36$, $y_2 = \frac{52-20}{2} = 16$ е) $15y^2 - 22y - 37 = 0$ $D = (-22)^2 - 4 \cdot 15 \cdot (-37) = 484 + 2220 = 2704 = 52^2$ $y_1 = \frac{22+52}{30} = 2,46... (\frac{74}{30} = \frac{37}{15})$, $y_2 = \frac{22-52}{30} = -1$ ж) $25p^2 - 10p + 1 = 0$ $D = (-10)^2 - 4 \cdot 25 \cdot 1 = 100 - 100 = 0$ $p = \frac{10}{50} = 0,2$ з) $400x^2 + 100x - 500 = 0 \Rightarrow 4x^2 + x - 5 = 0$ $D = 1^2 - 4 \cdot 4 \cdot (-5) = 1 + 80 = 81 = 9^2$ $x_1 = \frac{-1+9}{8} = 1$, $x_2 = \frac{-1-9}{8} = -1,25$