Помогите решить неравенство 2x^2-5x+2/x+4<0

$\frac{2 x^{2} -5x+2}{x+4} \ \textless \ 0 \\ 2 x^{2} -5x+2=0 \\ D=25-4*4=25-16=9 \\ \sqrt{D} =3 \\ x_{1} = \frac{5+3}{4} = \frac{8}{4} =2 \\ x_{2} = \frac{5-3}{4} = \frac{2}{4}= \frac{1}{2} =0.5 \\ x+4 \neq 0 \\ x \neq -4 \\ \\ \frac{2(x-0.5)(x-2)}{(x+4)} \ \textless \ 0 \\ \\$

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----------------|----------------------|------------------------|-----------------> x
-4 0.5 2

$x\in (- \infty} ;-4)\cup (0.5;2)$