$x\ge2\\bpt<=>x^2-3x+4\ge x+2 <=>x^2-4x+2 \ge 0 <=>x\ge 2+\sqrt2$ hoặc $x\le 2-\sqrt 2$vậy $S=[-2;2-\sqrt2]\cup [2;+\infty)$
$x\ge2\\bpt<=>x^2-3x+4\ge x+2
\\<=>x^2-4x+2 \ge 0
\\ $$<=>x\ge 2+\sqrt
2$ hoặc $x\le 2-\sqrt 2$vậy $S=[-2;2-\sqrt2]\cup [2;+\infty)$