541. Решите уравнение: a) 25 = 26x - x²; б) 3t² = 10 - 29t; в) у² = 4y + 96;

a) $$25 = 26x - x^2$$ $$x^2 - 26x + 25 = 0$$ $$D = (-26)^2 - 4 \cdot 1 \cdot 25 = 676 - 100 = 576$$ $$x_1 = \frac{26 + \sqrt{576}}{2 \cdot 1} = \frac{26 + 24}{2} = \frac{50}{2} = 25$$ $$x_2 = \frac{26 - \sqrt{576}}{2 \cdot 1} = \frac{26 - 24}{2} = \frac{2}{2} = 1$$ Ответ: $$x_1 = 25$$, $$x_2 = 1$$ б) $$3t^2 = 10 - 29t$$ $$3t^2 + 29t - 10 = 0$$ $$D = 29^2 - 4 \cdot 3 \cdot (-10) = 841 + 120 = 961$$ $$t_1 = \frac{-29 + \sqrt{961}}{2 \cdot 3} = \frac{-29 + 31}{6} = \frac{2}{6} = \frac{1}{3}$$ $$t_2 = \frac{-29 - \sqrt{961}}{2 \cdot 3} = \frac{-29 - 31}{6} = \frac{-60}{6} = -10$$ Ответ: $$t_1 = \frac{1}{3}$$, $$t_2 = -10$$ в) $$y^2 = 4y + 96$$ $$y^2 - 4y - 96 = 0$$ $$D = (-4)^2 - 4 \cdot 1 \cdot (-96) = 16 + 384 = 400$$ $$y_1 = \frac{4 + \sqrt{400}}{2 \cdot 1} = \frac{4 + 20}{2} = \frac{24}{2} = 12$$ $$y_2 = \frac{4 - \sqrt{400}}{2 \cdot 1} = \frac{4 - 20}{2} = \frac{-16}{2} = -8$$ Ответ: $$y_1 = 12$$, $$y_2 = -8$$