Решите неравенство -(3x+2)(2x-3)>0

0, (3x+2)(2x-3)<0, 6x^2-5x-6<0," alt="-(3x+2)(2x-3)>0, (3x+2)(2x-3)<0, 6x^2-5x-6<0," align="absmiddle" class="latex-formula">
$6x^2-5x-6=0, D=(-5)^2-4*6*(-6)=25+144=169=13^2,$
$x_{1} = \frac{5+13}{2*6} = \frac{3}{2} , x_{2} = \frac{5-13}{2*6} =- \frac{2}{3}$
$6(x- \frac{3}{2} )(x+ \frac{2}{3} )<0, (2x-3)(3x+2)<0$
Ответ: х∈ $[- \frac{2}{3} ; \frac{3}{2} ]$