Теңсіздік шешу а) (х+2)*(2-х)<3х^2-8 б) 2х-6<(3-х)*(х+3)

Решите задачу:

$a)(x+2)(2-x)\ \textless \ 3x^2-8 \\ \\ (2+x)(2-x)\ \textless \ 3x^2-8 \\ \\ 4-x^2\ \textless \ 3x^2-8 \\ \\ -x^2-3x^2\ \textless \ -8-4 \\ \\ -4x^2\ \textless \ -12 \\ \\ x^2\ \textgreater \ 3 \\ \\ (- \infty;- \sqrt{3} )U(\sqrt{3};+\infty) \\ \\ \\ b)2x-6\ \textless \ (3-x)(x+3) \\ \\ 2x-6\ \textless \ (3-x)(3+x) \\ \\ 2x-6\ \textless \ 9-x^2 \\ \\ 2x-6-9+x^2\ \textless \ 0 \\ \\ x^2+2x-15\ \textless \ 0 \\ \\ x_1=-5;x_2=3 \\ \\ (-5;3)$