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.

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