Решите неравенство ((3X^2+4x-4)/(8+15x))<0

$\frac{3x^{2}+4x-4}{8+15x}<0\\\\ ODZ:\\\\ 8+15x \neq 0\\ 15x \neq -8 \ | :15\\ x\neq-\frac{8}{15}\\\\ 3x^{2}+4x-4 = 0\\ D=b^{2}-4ac=4^{2}-4\cdot 3\cdot (-4)=16-(-48)=64\\ x_{1,2}=\frac{-b+/- \sqrt{D}}{2a}\\ x_{1}=\frac{-4+8}{2\cdot 3}=\frac{4}{6}=\frac{2}{3}\\ x_{2}=\frac{-4-8}{2\cdot 3}=\frac{-12}{6}=-2\\\\ \frac{3(x-\frac{2}{3})(x+2)}{8+15x}<0\\ x<\frac{2}{3}\\\\ x<-2$

/////////////-2////////////⅔------------------------------->x

x ∈ (-∞; -2) U (-2; ⅔)

Ответ: x ∈ (-∞; -2) U (-2; ⅔).