Решите относительно х уравнение

Решите задачу:

$\displaystyle (a - 1)x^2 + 2ax + a + 1 = 0\\\\ \text{A} = a - 1, \ \text{B} = 2a, \ \text{C} = a + 1;\\\\ 1) \ a \ne 1;\\\\ \text{D} = \text{B}^2 - 4\text{A}\text{C} = (2a)^2 - 4(a - 1)(a + 1) =\\\\= 4a^2 - 4(a^2 - 1) = 4a^2 - 4a^2 + 4 = 4\\\\ x_1 = \frac{-2a - 2}{2(a - 1)} = \frac{-2(a + 1)}{2(a - 1)} = -\frac{a + 1}{a - 1};\\\\ x_2 = \frac{-2a + 2}{2(a - 1)} = \frac{-2(a - 1)}{2(a - 1)} = -1\\\\ 2) \ a = 1; \\\\ 2x + 2 = 0, \ x = -1$