How many consecutive odd integers beginning with 5 will sum to 480?

How many consecutive odd integers beginning with 5 will sum to 480?
Solution :
Consecutive integers starting from 5 are 5, 7, 9, ............
5 + 7 + 9 + 11 + .............  = 480
Sn = 480
(n/2)[2a + (n - 1)d]  = 480
a  = 5, d = 7 - 5  = 2
(n/2)[2(5) + (n - 1)(2)]  = 480
(n/2)[10 + 2n - 2]  = 480
(n/2)[2n + 8]  = 480
2n2 + 8n  = 960
Dividing by 2, we get
n2 + 4n  - 480 =  0
(n + 24) (n - 20)  = 0
n = -24 or n = 20
Hence by finding the sum of 20 terms we will get 480.