Answer
$\frac{1+\sqrt{1+8N}}{2}$.
Work Step by Step
$\begin{align}
& N=\frac{1}{2}\left( {{n}^{2}}-n \right) \\
& =\frac{1}{2}{{n}^{2}}-\frac{1}{2}n
\end{align}$
Subtract $N$ on both sides of the equation,
$\begin{align}
& N-N=\frac{1}{2}{{n}^{2}}-\frac{1}{2}n-N \\
& 0=\frac{1}{2}{{n}^{2}}-\frac{1}{2}n-N
\end{align}$
Solving the equation for $n$,
$n=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substitute the values of the equation in the quadratic formula,
$a=\frac{1}{2},b=-\frac{1}{2},c=-N$
Then, the equation is,
$\begin{align}
& n=\frac{-\left( -\frac{1}{2} \right)\pm \sqrt{{{\left( -\frac{1}{2} \right)}^{2}}-4\times \frac{1}{2}\times \left( -N \right)}}{2\times \left( \frac{1}{2} \right)} \\
& =\frac{1}{2}\pm \sqrt{\frac{1}{4}+2N} \\
& =\frac{1}{2}\pm \frac{1}{2}\sqrt{1+8N} \\
& =\frac{1\pm \sqrt{1+8N}}{2}
\end{align}$
$n=\frac{1-\sqrt{1+8N}}{2},\text{ }\frac{1+\sqrt{1+8N}}{2}$
We only consider positive values.
$n=\frac{1+\sqrt{1+8N}}{2}$
Thus, the value is $n=\frac{1+\sqrt{1+8N}}{2}$.
