How do you solve for n in $S = \frac{n (n + 1)}{2}$ $S = (n(n+1))/ 2$ ?

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Answer 1 · 2016-06-07T07:18:46

$n = \frac{- 1 \pm \sqrt{1 + 8 S}}{2}$ $n = (-1 +- sqrt(1+8S))/(2)$

$S = \frac{n (n + 1)}{2}$ $S = (n(n+1))/2$

$S = \frac{n^{2} + n}{2}$ $S = (n^2+n)/2$

$2 S = n^{2} + n$ $2S = n^2 +n$

$n^{2} + n - 2 S = 0$ $n^2 + n - 2S = 0$

using the quadratic formula for : $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ ,

$n = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $n = (-b +- sqrt( b^2 - 4ac) ) /(2a)$

$n = \frac{- 1 \pm \sqrt{1 - 4 (1) (- 2 S)}}{2}$ $n = (-1 +- sqrt(1 - 4(1)(-2S)))/2$

$n = \frac{- 1 \pm \sqrt{1 + 8 S}}{2}$ $n = (-1 +- sqrt(1+8S))/(2)$

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