How do you solve for n in #S = (n(n+1))/ 2#?

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Answer 1 · 2016-06-07T07:18:46

# n = (-1 +- sqrt(1+8S))/(2) #

# S = (n(n+1))/2 #

# S = (n^2+n)/2 #

# 2S = n^2 +n #

# n^2 + n - 2S = 0#

using the quadratic formula for : # ax^2 + bx + c = 0#,

# n = (-b +- sqrt( b^2 - 4ac) ) /(2a) #

# n = (-1 +- sqrt(1 - 4(1)(-2S)))/2 #

# n = (-1 +- sqrt(1+8S))/(2) #