Помогите решить неравенство (2х-7)(3х+5) больше (x+9)(6x-13)

[latex](2x-7)(3x+5)\ \textgreater \ (x+9)(6x-13)[/latex] [latex]6x^2+10x-21x-35\ \textgreater \ 6x^2-13x+54x-117[/latex] [latex]6x^2-11x-35\ \textgreater \ 6x^2+41x-117[/latex] [latex]6x^2-11x-35-6x^2-41x+117\ \textgreater \ 0[/latex] [latex]-52x+82\ \textgreater \ 0[/latex] [latex]-52x\ \textgreater \ -82[/latex]   [latex]26x\ \textless \ 41[/latex] [latex]x\ \textless \ \frac{41}{26} [/latex] [latex]x\ \textless \ 1 \frac{15}{26} [/latex] Ответ: [latex](-[/latex]∞[latex];1 \frac{15}{26} )[/latex]