Решите неравенство

Решите задачу:

$\frac{6}{ {x}^{2} - x - 6} < - 1 \\ odz \\ {x}^{2} - x - 6≠0 \\ d = 1 + 4 \times 6 = 25 = {5}^{2} \\ x ≠ \frac{1±5}{2} = 3;- 2 \\ \frac{6}{ {x}^{2} - x - 6 } + 1 < 0 \\ \frac{6 + {x}^{2} - x - 6}{ {x}^{2} - x - 6} \\ \frac{ {x }^{2} - x}{ {x}^{2} - x - 6 } < 0 \\ {x}^{2} - x = 0 \\ x(x - 1) = 0 \\ x = 0 \\ x = 1 \\ + + ( - 2) - - (0 )+ + (1) - - (3 ) + + \\ x\in( - 2;0)\cup(1;3)$