Wendy, Elizabeth, and Charlie are all working on a math problem together and they are having a disagreement.: Ticket lines are huge at the Math Olympics ticket office. Pi, the local math team, is competing to be the Math Champions. When tickets went on sale, 500 were sold in the first hour alone. In the second hour 560 tickets were sold. After the second hour, the increase in ticket sales every hour was 20 more than the increase of the previous hour. Write an equation to model the number of ticket sales each hour for this event. If sales continue at this rate, in what hour will 11,000 tickets be sold for the championship match? Is this quadratic, linear, exponential, or something else? If it is, how do you know? If it is something else, why?
Wendy thinks this is a quadratic equation. Charlie thinks this is exponential. Elizabeth thinks it is something else. Who is right? Explain who is right, write an equation to model the number of ticket sales each hour for this event, and answer the following question: If sales continue at this rate, in what hour will 11,000 tickets be sold for the championship match?
Included was this table to get us started. I added what I could figure out in green. 
I could figure out the y-intercept and the other values because I know the second difference. At first I thought it was quadratic, since I know if the second difference is constant, it is quadratic. I was able to write an equation, (if x=hour and y=#of tickets sold in that hour): y=10 x^2+30 x +460. However, I couldn't figure out how to find what I was asked, since I realized they wanted to know on what hour is the first difference +11,000, not when y=11,000. Now I'm starting to think it is something else, like a sequence. I am stuck here. Am I even on the the right track? If som where do I go next? If not, what am I supposed to do?
If the second difference is a constant, the function must be a quadratic. So, let us write $$y(t)=a+b\,t+c\,t^2$$ and let us apply what we already know $$y(1)=a+b+c=500$$ $$y(2)=a+2 b+4 c=560$$ $$(y(3)-y(2))-(y(2)-y(1))=2c=20$$ So three linear equations in $a,b,c$, easy to solve by elimination, leading to $a=460$, $b=30$, $c=10$ (what you obviously obtained).
So, your green numbers are perfectly correct.
Last question : at what time will $11000$ tickets be solved ? This is given by $$460+30t+10t^2=11000$$ which is a simple quadratic equation; its roots are $-34$ (to be discarded since negative) and $31$ which is the solution.
Is this clear to you ?
Edit
As you properly thought, you could have been considering series writing the second difference as $$u_{n+2}-2u_{n+1}+u_n=20$$ the general solution of which being $$u_n=c_1+c_2 n+10 \left(n^2-n\right)$$ and apply the conditions $u_1=500$, $u_2=560$ to get the same result.
From this $u_{n+1}-u_n=40+20n$ and $11000$ tickets will be sold during the $548^{th}$ hour.