Помогите пожалуйста!!!

$\frac{3}{ {x}^{2} - 1} - \frac{1}{2x + 2} = \frac{1}{4} \\ \frac{3}{(x - 1)( x + 1)} - \frac{1}{2(x + 1)} - \frac{1}{4} = 0 \\ \frac{12}{4(x - 1)(x + 1)} - \frac{2(x - 1)}{4(x - 1)(x + 1)} - \frac{(x - 1)(x + 1)}{4(x - 1)(x + 1)} = 0 \\ \frac{12 - 2(x - 1) - (x - 1)(x + 1)}{4(x - 1)(x + 1)} = 0 \\ \frac{12 - 2x + 2 - {x}^{2} + 1}{4(x - 1)(x + 1)} = 0 \\ \frac{ - {x}^{2} - 2x + 15 }{4(x - 1)(x + 1)} = 0 \\ - {x}^{2} - 2x + 15 = 0 \\ {x}^{2} + 2x - 15 = 0 \\ d = {b}^{2} - 4ac = 4 - 4 \times ( - 15) = 64 \\ x1 = \frac{ - 2 + 8}{2} = \frac{6}{2} = 3 \\ x2 = \frac{ - 2 - 8}{2} = \frac{ - 10}{2} = - 5$
Ответ: -5; 3.