Решите уравнение

ОДЗ:
x² - 1 ≠ 0
x ≠ -1; 1
x² - 6x + 8 = (2x² - 6x - 3)/(x² - 1)
x² - 6x + 9 - 1 = (2x² - 6x - 3)/(x² - 1)
(x - 3)² - 1 = (2x² - 6x - 3)/(x - 1)(x + 1)
(x - 3 - 1)(x - 3 + 1) = (2x² - 6x - 3)/(x - 1)(x + 1)
(x - 4)(x - 2)(x - 1)(x + 1) = (2x² - 6x - 3)
(x² + x - 4x - 4)(x² - x - 2x + 2) = (2x² - 6x - 3)
(x² - 3x - 4)(x² - 3x + 2) = (2(x² - 3) - 3))
Пусть t = (x² - 3x).
(t - 4)(t + 2) = 2t - 3
t² + 2t - 4t - 8 = 2t - 3
t² - 2t - 8 = 2t - 3
t² - 4t - 5= 0
t² - 4t + 4 - 9 = 0
(t - 2)² - 3² = 0
(t - 2 - 3)(t - 2 + 3) = 0
(t - 5)(t + 1) = 0
t = -1; 5.
Обратная замена:
1) t = -1
x² - 3x = -1
x² - 3x + 1 = 0
D = 9 - 4 = 5
$x_1 = \dfrac{3 + \sqrt{5} }{2} \\ \\ x_2 = \dfrac{3 - \sqrt{5} }{2}.$
2) t = 5
x² - 3x = 5
x² - 3x - 5 = 0
D = 9 + 20 = 29
$x_1 = \dfrac{3 + \sqrt{29} }{2} \\ \\ x_2 = \dfrac{3 - \sqrt{29} }{2}$
Ответ: x = (3 ± √5)/2; (3 ± √29)/2.