User:Wguynes/Proofs/Triangular number test

Let T = number to test

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

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

Find the roots of the quadratic. What value of n would make our T?

$n={\frac {-1\pm {\sqrt {1^{2}-4(1)(-2T)}}}{2\cdot 1}}={\frac {-1\pm {\sqrt {8T+1}}}{2}}$

Only one of these two roots can be a positive result.

$n={\frac {{\sqrt {8T+1}}-1}{2}}$