Решите пожалуйста уравнение а)Интеграл от 1 до Х(1-4t)dt=12-9x b)Интеграл от Х до...

A)
$\int\limits^{x}_{1} {(1-4t)} \, dt = t - 2t^2 + C \bigg|_{1}^{x} = x - 2x^2 - 1 + 2 = 12-9x;\\\\\\ -2x^2 + x + 1 +9x - 12 = 0;\\ -2x^2 + 10x - 11 = 0;\\ D = 100 - 88 = 12\\ x_1 = -\frac{-10 + \sqrt{12}}{4}\\ x_2 = -\frac{-10 - \sqrt{12}}{4}\\$

b)
$\int\limits^{-1}_{x} {(3t-2)} \, dt = \frac{3}{2}t^2 - 2t + C \bigg|^{-1}_{x} = \frac{3}{2} + 2 - \frac{3}{2}x^2 + 2x = - \frac{3}{2}x^2 + 2x + 3,5 = 5-x;\\\\\\-\frac{3}{2}x^2 + 2x + 3,5 = 5-x;\\ - \frac{3}{2}x^2 + 3x - 1,5 = 0;\\ D = 9 - 9 = 0\\ x = \frac{-3}{2\cdot\frac{-3}{2}} = 1$

c)
$\int\limits^{-2}_{x} {(5t+1)} \, dt = 2,5t^2 + t + C \bigg|^{-2}_{x} = 10 + 2 - 2,5x^2 - x = -2,5x^2 - x + 12 = 6+x;\\\\\\-2,5x^2 - 2x + 6 = 0;\\ D = 4 + 60 = 64 = (\pm{8})^2\\ x_1 = -\frac{2 + 8}{5} = -2\\ x_2 = -\frac{2 - 8}{5} = \frac{6}{5}\\$