$(x^2 - 4x)^2 - 2(x^2 - 4x) - 15 = 0$
$y = x^2 - 4x$
$y^2 - 2y - 15 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * (-15) = 4 + 60 = 64 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{64}}{2 * 1} = \frac{2 + 8}{2} = \frac{10}{2} = 5$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{64}}{2 * 1} = \frac{2 - 8}{2} = \frac{-6}{2} = -3$
$x^2 - 4x = 5$
$x^2 - 4x - 5 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * (-5) = 16 + 20 = 36 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{36}}{2 * 1} = \frac{4 + 6}{2} = \frac{10}{2} = 5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{36}}{2 * 1} = \frac{4 - 6}{2} = \frac{-2}{2} = -1$
или
$x^2 - 4x = -3$
$x^2 - 4x + 3 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * 3 = 16 - 12 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{4}}{2 * 1} = \frac{4 + 2}{2} = \frac{6}{2} = 3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{4}}{2 * 1} = \frac{4 - 2}{2} = \frac{2}{2} = 1$
Ответ:
А) −1; 1; 3; 5