Precalculus, Chapter 9, 9.4, Section 9.4, Problem 72

You need to remember what a quadratic model is, such that:
a_n = f(n) = a*n^2 + b*n + c
The problem provides the following information, such that:
a_0 = 3 => f(0) = a*0^2 + b*0 + c => c = 3
a_2 = 0 => f(2) = a*2^2 + b*2 + c => 4a + 2b + c = 0
a_6 = 36 => f(6) = a*6^2 + b*6 + c => 36a + 6b + c = 36
You need to replace 3 for c in the next two equations, such that:
4a + 2b + 3 = 0 => 4a + 2b = -3
36a + 6b + 3 = 36 => 36a + 6b = 33 => 18a + 2b = 11
Subtract the equation 4a + 2b = -3 from the equation 18a + 2b = 11 :
18a + 2b - 4a - 2b = 11 + 3
14a = 14 => a = 1
Replace 1 for a in equation 4a + 2b = -3 , such that:
4 + 2b = -3 => 2b = -7 => b = -7/2
Hence, the quadratic model for the given sequence is a_n = n^2 - (7/2)*n + 3.