Motivation:
Let $n,e,d$ be integers greater thant 2, such that $e\mid n-1$ and $d\mid n-1$. For $i=1,2,3,4,5$, define $$L_i=L_i(n,e,d)=a_{i1}n+a_{i2}e+a_{i3}d,$$ and $$k=\lceil \frac{L_1}{L_2} \rceil.$$ Suppose $$L_3 k^2 + L_4 k + L_5<0.$$ Denote $\delta=e-d$, i.e. $d=e+\delta$. Find out $G_i=G_i(n,e)$, $i=1,2$, such that $G_1 < \delta < G_2$.
My trying: If $k=\frac{L_1}{L_2}$, then I would solve the problem as follows.
1) Multiplying both sides by $L_2^2$, $L_3 k^2 + L_4 k + L_5<0$ if and only if $L_3L_1^2+L_4L_1L_2+L_5L_2^2<0$
2) Replace $d$ with $e+\delta$. We have four functions $H_i=H_i(n,e)$, $i=1,2,3,4$ such that $$H_1 \delta^3+ H_2 \delta^2 + H_3 \delta + H_4 < 0.$$
3) Solve the cubic inquality to obtain $G_1$ and $G_2$.
However it does not work in the first step if there exists ceiling function. Of course, by the ceiling function we have $$\frac{L_1}{L_2}\le k < \frac{L_1}{L_2}+1.$$ Denote $LHS=L_3 k^2 + L_4 k + L_5$. For simplicity, assume $L_2>0$, then $P_1\le LHS < P_2,$ where $$P_1=\frac{L_3L_1^2+L_4L_1L_2+L_5L_2^2}{L_2^2},$$ and $$P_2=\frac{L_3(L_1^2+2L_1L_2+L_2^2)+L_4(L_1L_2+L_2^2)+L_5L_2^2}{L_2^2}.$$ I am not sure if the sufficient and necessary condition of $LHS<0$ is $P2<=0$. However my question is a bit complicated.
3.Question
Let $n,e,d$ be integers greater thant 2, such that $e\mid n-1$ and $d\mid n-1$. Define $$L_1=L_1(n,d)=dn+d+n-1,$$ $$L_2=L_2(n,e,d)=e(d+n-1)$$ and $$k=\lceil \frac{L_1}{L_2} \rceil.$$ What is the sufficient and necessary condition of $$L_1 k^2 -(2L_2+L_1) k + (e+1)L_2-d(2en+e-3)<0,$$ having the form $G_1 < \delta < G_2$, where $\delta=e-d$ and $G_i=G_i(n,e)$, $i=1,2$?