$(x^2 - 2)^2 - 8(x^2 - 2) + 7 = 0$
$y = x^2 - 2$
$y^2 - 8y + 7 = 0$
$D = b^2 - 4ac = (-8)^2 - 4 * 1 * 7 = 64 - 28 = 36 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{8 + \sqrt{36}}{2 * 1} = \frac{8 + 6}{2} = \frac{14}{2} = 7$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{8 - \sqrt{36}}{2 * 1} = \frac{8 - 6}{2} = \frac{2}{2} = 1$
$x^2 - 2 = 7$
$x^2 = 7 + 2$
$x^2 = 9$
x = ±3
или
$x^2 - 2 = 1$
$x^2 = 1 + 2$
$x^2 = 3$
$x = ±\sqrt{3}$
Ответ: $-3, -\sqrt{3}, \sqrt{3}, 3.$

$(x^2 + 5x)^2 - 2(x^2 + 5x) - 24 = 0$
$y = x^2 + 5x$
$y^2 - 2y - 24 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * (-24) = 4 + 96 = 100 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{100}}{2 * 1} = \frac{2 + 10}{2} = \frac{12}{2} = 6$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{100}}{2 * 1} = \frac{2 - 10}{2} = \frac{-8}{2} = -4$
$x^2 + 5x = 6$
$x^2 + 5x - 6 = 0$
$D = b^2 - 4ac = 5^2 - 4 * 1 * (-6) = 25 + 24 = 49 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{49}}{2 * 1} = \frac{-5 + 7}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{49}}{2 * 1} = \frac{-5 - 7}{2} = \frac{-12}{2} = -6$
или
$x^2 + 5x = -4$
$x^2 + 5x + 4 = 0$
$D = b^2 - 4ac = 5^2 - 4 * 1 * 4 = 25 - 16 = 9 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{9}}{2 * 1} = \frac{-5 + 3}{2} = \frac{-2}{2} = -1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{9}}{2 * 1} = \frac{-5 - 3}{2} = \frac{-8}{2} = -4$
Ответ: −6, −4, −1, 1.

$(x^2 - 3x + 1)(x^2 - 3x + 3) = 3$
$y = x^2 - 3x$
(y + 1)(y + 3) = 3
$y^2 + y + 3y + 3 - 3 = 0$
$y^2 + 4y = 0$
y(y + 4) = 0
y = 0
или
y + 4 = 0
y = −4
$x^2 - 3x = 0$
x(x − 3) = 0
x = 0
или
x − 3 = 0
x = 3
или
$x^2 - 3x = -4$
$x^2 - 3x + 4 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * 4 = 9 - 16 = -7 < 0$ − нет корней
Ответ: 0 и 3

$(x^2 + 2x + 2)(x^2 + 2x - 4) = -5$
$y = x^2 + 2x$
(y + 2)(y − 4) = −5
$y^2 + 2y - 4y - 8 + 5 = 0$
$y^2 - 2y - 3 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * (-3) = 4 + 12 = 16 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{16}}{2 * 1} = \frac{2 + 4}{2} = \frac{6}{2} = 3$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{16}}{2 * 1} = \frac{2 - 4}{2} = \frac{-2}{2} = -1$
$x^2 + 2x = 3$
$x^2 + 2x - 3 = 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-3) = 4 + 12 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{16}}{2 * 1} = \frac{-2 + 4}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{16}}{2 * 1} = \frac{-2 - 4}{2} = \frac{-6}{2} = -3$
или
$x^2 + 2x = -1$
$x^2 + 2x + 1 = 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * 1 = 4 - 4 = 0$
$x = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{0}}{2 * 1} = \frac{-2}{2} = -1$
Ответ: −3, −1, 1.