Решите неравенство втором способом!!!!!! номер 5.33 (а)

Решите задачу:

$|x^2-x-2| \leq -2x-2\\\\2x+2 \leq x^2-x-2 \leq -2x-2\\\\ \left \{ {{x^2-x-2 \leq -2x-2} \atop {x^2-x-2 \geq 2x+2}} \right. \; ;\; \left \{ {{x^2+x \leq 0} \atop {x^2-3x-4 \geq 0}} \right. \; ;\; \left \{ {{x(x+1) \leq 0} \atop {(x+1)(x-4) \geq 0}} \; ;\; \left \{ {{x\in [-1;0\, ]} \atop {x\in (-\infty ;-1\, ]\cup [\, 4\; +\infty )}} \right. \right. \\\\-----(-1)//////(0)---------\\//////////(-1)-------(4)/////////\\\\x=-1\\\\$