The sum of the squares of 3 consecutive positive integers is 116. what are the numbers?

which of the following equations is used in the process of solving this problem?

a. 3n^2 + 5 = 116
b. 3n^2 + 3n + 3 = 116
c. 3n^2 + 6n + 5 = 116

Let's say, the first number is "n"

well, the consecutive number of "n" is "n + 1"and the consecutive number of "n+1" is "n + 1" +1 or n + 2

so, the numbers are (n), (n+1) and (n+2), whatever "n" is

now, the sum of their squares is 116



expand the binomials by FOIL or binomial theorem, and then simplify