smc

Set the two equations equal to each other: $x^2+ax+b=x^2+cx+d$. Now remove the x squared and get $x$'s on one side: $ax-cx=d-b$. Now factor $x$: $x(a-c)=d-b$. If $a$ cannot equal $c$, then there is always a solution, but if $a=c$, a $1$ in $6$ chance, leaving a $1080$ out $1296$, always having at least one point in common. And if $a=c$, then the only way for that to work, is if $d=b$, a $1$ in $36$ chance, however, this can occur $6$ ways, so a $1$ in $6$ chance of this happening. So adding one thirty sixth to $\frac{1080}{1296}$, we get the simplified fraction of $\frac{31}{36}$; answer $\boxed{\frac{31}{36}}$.