Unpaired electrons from magnetic moment


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> A complex formulated as $\ce{Cs\_x[NbCl6]}$ has spin-only magnetic moment $\mu\_{s.o}$ of approximately 2 **B.M**. From this value calculate how many unpaired electrons are present.
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Spin-only magnetic moment is related to the number of unpaired electrons by the formula 


$$\mu\_{s.o}= \sqrt{n(n+1)}$$ where $n$ is the number of unpaired electrons. 


I let $\sqrt{n(n+1} = 2$ and solved for $n$. I got two values one $+$ and one $-$. Do I just take the positive value and round it to the nearest whole number. In this case I got $n\_+= \sqrt{5} - 1 \approx 1.2$. 


Does this mean there is $1$ unpaired electron?


