The variance is $$\sigma_y^2=\frac{1}{n-1}\sum_{i=1}^n(y_i-\bar{y})^2$$ and I am told that we can write it as $$\frac{1}{n-1}\left(\sum_{I=1}^ny_i^2-n\bar{y}^2\right)$$
My attempt
$$\sigma_y^2=\frac{1}{n-1}\sum_{i=1}^n(y_i-\bar{y})^2 = \frac{1}{n-1}\left(\sum_{i=1}^n(y_i^2+\bar{y}^2-2y_i\bar{y})\right) = \frac{1}{n-1}\left(\sum_{i=1}^ny_i^2+n\bar{y}^2-2\bar{y}\sum_{i=1}^ny_i\right)$$
How do I continue??
Hint: You are almost done ... Note, that $$\sum_{i=1}^{n}y_i=n\bar{y}.$$