Решите неравенство: (3x^2+5x-12)√16-x^2< =0

ОДЗ
16-x²≥0
(4-x)(4+x)≥0
x=4 U x=-4
x∈[-4;4]
√(16-x²)≥0⇒3x²+5x-12≤0
D=25+144=169
x1=(-5-13)/6=-3
x2=(-5+13)/6=4/3
-3≤x≤4/3
x∈[-3;1 1/3] U {-4;4}