Помогите решить,пожалуйста:)

Решите задачу:

$2-\dfrac{104}{12n-5}\ \textless \ 0\\ 2(12n-5)^2-104(12n-5)\ \textless \ 0\\ 2(144n^2-120n+25)-1248n+520\ \textless \ 0\\ 288n^2-240n+50-1248n+520\ \textless \ 0\\ 288n^2-1488n+570\ \textless \ 0\\ 48n^2-248n+95\ \textless \ 0\\ \Delta=(-248)^2-4\cdot48\cdot95=61504-18240=43264\\ \sqr{\Delta}=\sqrt{43264}=208\\ n_1=\dfrac{-(-248)-208}{2\cdot48}=\dfrac{40}{96}=\dfrac{5}{12}\\ n_2=\dfrac{-(-248)+208}{2\cdot48}=\dfrac{456}{96}=\dfrac{19}{4}\\ n\in\left(\dfrac{5}{12},\dfrac{19}{4}\right)\\\\ n\in\left(\dfrac{5}{12},\dfrac{19}{4}\right) \wedge n\in \mathbb{N}\\ \boxed{n=\{1,2,3,4\}}$