Помогите решить систему неравенств

$\left \{ {{x^2 - 8x - 12 \leq -5x - 8} \atop {-x^2+14x-18 \leq 4x+6}} \right. x^2 - 8x - 12 \leq -5x - 8 x^2 - 8x -12 + 5x + 8 \leq 0 x^2 - 3x - 4 = 0 D = 25 \left \{ {{x1 = 4} \atop {x2=-1}} \right.$

x ∈ [ - 1 ; 4 ]

$-x^2 + 14x - 18 \leq 4x+6 -x^2 + 14x - 18 - 4x - 6 \leq 0 -x^2 + 10x - 24 = 0 D = 4 \left \{ {{x1 = 4} \atop {x2=6}} \right.$

x ∈ ( - ∞ ; 4 ] U [ 6 ; + ∞ )

Итог:
$-1 \leq x \leq 4$