$(3x - 1)^4 - 20(3x - 1)^2 + 64 = 0$
$((3x - 1)^2)^2 - 20(3x - 1)^2 + 64 = 0$
$(3x - 1)^2 = y$, y ≥ 0
$y^2 - 20y + 64 = 0$
$D = b^2 - 4ac = (-20)^2 - 4 * 1 * 64 = 400 - 256 = 144 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{20 + \sqrt{144}}{2 * 1} = \frac{20 + 12}{2} = \frac{32}{2} = 16$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{20 - \sqrt{144}}{2 * 1} = \frac{20 - 12}{2} = \frac{8}{2} = 4$
$(3x - 1)^2 = 16$
3x − 1 = 4
3x = 4 + 1
3x = 5
$x = \frac{5}{3}$
$x = 1\frac{2}{3}$
или
3x − 1 = −4
3x = −4 + 1
3x = −3
x = −1

$(3x - 1)^2 = 4$
$3x - 1 = 2$
$3x = 2 + 1$
3x = 3
x = 1
или
$3x - 1 = -2$
$3x = -2 + 1$
$3x = -1$
$x = -\frac{1}{3}$
Ответ: $-\frac{1}{3}$, −1, 1, $1\frac{2}{3}$.

$(2x + 3)^4 - 24(2x + 3)^2 - 25 = 0$
$((2x + 3)^2)^2 - 24(2x + 3)^2 - 25 = 0$
$(2x + 3)^2 = y$, y ≥ 0
$y^2 - 24y - 25 = 0$
$D = b^2 - 4ac = (-24)^2 - 4 * 1 * (-25) = 576 + 100 = 676 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{24 + \sqrt{676}}{2 * 1} = \frac{24 + 26}{2} = \frac{50}{2} = 25$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{24 - \sqrt{676}}{2 * 1} = \frac{24 - 26}{2} = \frac{-2}{2} = -1$ − не удовлетворяет условию, так как y ≥ 0.
$(2x + 3)^2 = 25$
2x + 3 = 5
2x = 5 − 3
2x = 2
x = 1
или
2x + 3 = −5
2x = −5 − 3
2x = −8
x = −4
Ответ: −4 и 1