What is the maximum amount of solutions to $f(x+1)f(x)= ax^2+bx+c$.

Question

$f(x)$ is going to be in the form $mx+h$ thus, $(mx+m+h)(mx+h) = ax^2+bx+c$. With basic algebra $m= \pm \sqrt{a}$. Also $(m+h)(h)=c$. I would guess that because $(m+h)h=c$ has two solutions max if $m$ is one number, there will be 4 solutions max if $m$ is two numbers am I right? I'm trying to find the max amount of solutions to $f(x)f(x+1)= ax^2+bx+c$