\(a_{n-1} = a(n-1)^2 + b(n-1) + c = an^2 - 2an + a + bn - b + c\text{.}\) Therefore \(a_n - a_{n-1} = 2an - a + b\text{,}\) which is arithmetic. Notice that this is not quite the derivative of \(a_n\text{,}\) which would be \(2an + b\text{,}\) but it is close.
in-context