(3х-1)(4+х) больше 4 (4х-1) решить неравенство

[latex](3x-1)(4+x)\ \textgreater \ 4(4x-1) \\ 12x+3 x^{2} -4-x\ \textgreater \ 16x-4 \\ 3 x^{2} +12x-x-4\ \textgreater \ 16x-4 \\ 3 x^{2} +11x-4-16x+4\ \textgreater \ 0 \\ 3x^{2} +11x-16x\ \textgreater \ 0 \\ 3x^{2} -5\ \textgreater \ 0 \\ 3 x^{2} \ \textgreater \ 5 \\ x\ \textgreater \ \frac{5}{3} } \\ x\ \textless \ - \frac{5}{3} \\ x\in (- \infty} ;- \frac{5}{3} )\cup ( \frac{5}{3} ;+ \infty} )[/latex]