Кто сможет

$8x-12 \leq 2 x^{2} -2x$
$8x-12-2 x^{2} -2x \leq 0$
$10x-12-2x^{2} \leq 0$
$5x-6-x^{2} \leq 0$
$- x^{2} +5x-6 \leq 0$
$- x^{2} +3x+2x-6 \leq 0$
$-x(x-3)+2(x-3) \leq 0$
$(-x+2)(x-3) \leq 0$
рассматриваем два случая:
$1) \left \{ {-x+2 \leq 0} \atop {x-3 \geq 0}} \right. \left \{ {{x \geq 2} \atop {x \geq 3}} \right.$
$2) \left \{ {{-x+2 \geq 0} \atop {x-3 \leq 0}} \right. \left \{ {{x \leq 2} \atop {x \leq 3}} \right.$
1) x∈[3,+∞)
2) x∈(-∞,2]
ответ: x∈(-∞,2]∪[3,+∞)