Let $\sigma$ be variance and $\mu$ be mean
$\sigma = \sum_{i=1}^{10} \frac{(x_i - \mu)^2}{n} \\= \sum_{i=1}^{10} \frac{{x_i}^2}{n} - \mu^2 \\= \frac{0^2 + (-1)^2 + \dots + (-9)^2}{10} - (-4.5)^2 \\= \frac{ 9 . 10 . 19} {6 . 10} - 20.25\\ \left(\because \text{Sum of squares of first $n$ natural numbers $= \frac{n.(n+1).(2n+1)}{6}$} \right) \\= 28.5 - 20.25 = 8.25$