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

$(2x-7)(3x+5)\ \textgreater \ (x+9)(6x-13)$

$6x^2+10x-21x-35\ \textgreater \ 6x^2-13x+54x-117$

$6x^2-11x-35\ \textgreater \ 6x^2+41x-117$

$6x^2-11x-35-6x^2-41x+117\ \textgreater \ 0$

$-52x+82\ \textgreater \ 0$

$-52x\ \textgreater \ -82$

$26x\ \textless \ 41$

$x\ \textless \ \frac{41}{26}$

$x\ \textless \ 1 \frac{15}{26}$

Ответ: $(-$ ∞ $;1 \frac{15}{26} )$