$|x^2 - 2x - 6| = 6$
а)
$x^2 - 2x - 6 = 6$
$x^2 - 2x - 6 - 6 = 0$
$x^2 - 2x - 12 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * (-12) = 4 + 48 = 52 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{52}}{2 * 1} = \frac{2 + \sqrt{4 * 13}}{2} = \frac{2 + 2\sqrt{13}}{2} = \frac{2(1 + \sqrt{13})}{2} = 1 + \sqrt{13}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{52}}{2 * 1} = \frac{2 - \sqrt{4 * 13}}{2} = \frac{2 - 2\sqrt{13}}{2} = \frac{2(1 - \sqrt{13})}{2} = 1 - \sqrt{13}$
б)
$x^2 - 2x - 6 = -6$
$x^2 - 2x - 6 + 6 = 0$
$x^2 - 2x = 0$
x(x − 2) = 0
x = 0
или
x − 2 = 0
x = 2
Ответ: $1 - \sqrt{13}$; $1 + \sqrt{13}$; 0; 2.

$x^2 - 6|x| - 16 = 0$
а) x ≥ 0:
$x^2 - 6x - 16 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 1 * (-16) = 36 + 64 = 100 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{100}}{2 * 1} = \frac{6 + 10}{2} = \frac{16}{2} = 8$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{100}}{2 * 1} = \frac{6 - 10}{2} = \frac{-4}{2} = -2$ − не удовлетворяет условию, так как x ≥ 0.
б) x < 0:
$x^2 + 6x - 16 = 0$
$D = b^2 - 4ac = 6^2 - 4 * 1 * (-16) = 36 + 64 = 100 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-6 + \sqrt{100}}{2 * 1} = \frac{-6 + 10}{2} = \frac{4}{2} = 2$ − не удовлетворяет условию, так как x < 0.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-6 - \sqrt{100}}{2 * 1} = \frac{-6 - 10}{2} = \frac{-16}{2} = -8$
Ответ: −8 и 8

$x|x| + 2x - 15 = 0$
а) x ≥ 0:
$x^2 + 2x - 15 = 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-15) = 4 + 60 = 64 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{64}}{2 * 1} = \frac{-2 + 8}{2} = \frac{6}{2} = 3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{64}}{2 * 1} = \frac{-2 - 8}{2} = \frac{-10}{2} = -5$ − не удовлетворяет условию, так как x ≥ 0.
б) x < 0:
$-x^2 + 2x - 15 = 0$ | * (−1)
$x^2 - 2x + 15 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * 15 = 4 - 60 = -56 < 0$
нет корней
Ответ: 3

$||x^2 - 6x - 4| - 3| = 1$
1 случай
$|x^2 - 6x - 4| - 3 = 1$
$|x^2 - 6x - 4| = 1 + 3$
$|x^2 - 6x - 4| = 4$
а)
$x^2 - 6x - 4 = 4$
$x^2 - 6x - 4 - 4 = 0$
$x^2 - 6x - 8 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 1 * (-8) = 36 + 32 = 68 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{68}}{2 * 1} = \frac{6 + \sqrt{4 * 17}}{2} = \frac{6 + 2\sqrt{17}}{2} = \frac{2(3 + \sqrt{17})}{2} = 3 + \sqrt{17}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{68}}{2 * 1} = \frac{6 - \sqrt{4 * 17}}{2} = \frac{6 - 2\sqrt{17}}{2} = \frac{2(3 - \sqrt{17})}{2} = 3 - \sqrt{17}$
б)
$x^2 - 6x - 4 = -4$
$x^2 - 6x - 4 + 4 = 0$
$x^2 - 6x = 0$
x(x − 6) = 0
x = 0
или
x − 6 = 0
x = 6
2 случай
$|x^2 - 6x - 4| - 3 = -1$
$|x^2 - 6x - 4| = -1 + 3$
$|x^2 - 6x - 4| = 2$
а)
$x^2 - 6x - 4 = 2$
$x^2 - 6x - 4 - 2 = 0$
$x^2 - 6x - 6 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 1 * (-6) = 36 + 24 = 60 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{68}}{2 * 1} = \frac{6 + \sqrt{4 * 17}}{2} = \frac{6 + 2\sqrt{17}}{2} = \frac{2(3 + \sqrt{17})}{2} = 3 + \sqrt{17}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{68}}{2 * 1} = \frac{6 - \sqrt{4 * 17}}{2} = \frac{6 - 2\sqrt{17}}{2} = \frac{2(3 - \sqrt{17})}{2} = 3 - \sqrt{17}$
б)
$x^2 - 6x - 4 = -2$
$x^2 - 6x - 4 + 2 = 0$
$x^2 - 6x - 2 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 1 * (-2) = 36 + 8 = 44 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{44}}{2 * 1} = \frac{6 + \sqrt{4 * 11}}{2} = \frac{6 + 2\sqrt{11}}{2} = \frac{2(3 + \sqrt{11})}{2} = 3 + \sqrt{11}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{44}}{2 * 1} = \frac{6 - \sqrt{4 * 11}}{2} = \frac{6 - 2\sqrt{11}}{2} = \frac{2(3 - \sqrt{11})}{2} = 3 - \sqrt{11}$
Ответ: $3 + \sqrt{17}$; $3 - \sqrt{17}$; 0; 6; $3 + \sqrt{15}$; $3 - \sqrt{15}$; $3 + \sqrt{11}$; $3 - \sqrt{11}$.