Precalculus, Chapter 9, 9.4, Section 9.4, Problem 74

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 = -5 => f(2) = a*2^2 + b*2 + c => 4a + 2b + c = -5
a_6 = -57 => f(6) = a*6^2 + b*6 + c => 36a + 6b + c = -57
You need to replace -3 for c in equation 4a + 2b + c = -5 :
4a + 2b - 3 = -5 => 4a + 2b = -2 => 2a + b = -1
You need to replace -3 for c in equation 36a + 6b + c = -57 :
36a + 6b - 3 = -57 => 36a + 6b = -54 => 6a + b = -9
Subtract 2a + b = -1 from 6a + b = -9 , such that:
6a + b - 2a - b= -9 + 1
4a = -8 => a = -2
Replace -2 for a in equation 2a + b = -1 such that:
2*(-2) + b = -1=>-4 + b = -1 => b = 3
Hence, the quadratic model for the given sequence is a_n = -2n^2 + 3n - 3.