Assuming that $\bar{x} = \frac{\sum_{i = 1}^n x_i}{n}$, I have the following expression that I'm trying to simplify:
$\sum_{i = 1}^n (x_i - \bar{x})^2 + n\bar{x}^2$
$ = \sum_{i = 1}^n (x_i^2 - 2\bar{x}x_i + \bar{x}^2) + n\bar{x}^2$
$ = \sum_{i = 1}^n (x_i^2) - 2n\bar{x}\sum_{i = 1}^n x_i + n\bar{x}^2 + n\bar{x}^2$
$ = \sum_{i = 1}^n (x_i^2) - 2(\sum_{i = 1}^n x_i)^2 + \frac{2(\sum_{i = 1}^n x_i)^2}{n}$
However, the solution is
$ = \sum_{i = 1}^n x_i^2$?
How did $-2(\sum_{i = 1}^n x_i)^2 + \frac{2(\sum_{i = 1}^n x_i)^2}{n} = 0$?
As lulu commented, your mistake was made at the second equal sign. It should be: $$=\sum_{i=1}^n x_i^2 - 2\bar x \sum_{i=1}^n x_i + n\bar{x}^2 + n\bar{x}^2 = \sum_{i=1}^n x_i^2 -2\bar{x}(n\bar{x}) + 2n\bar{x}^2 = \sum_{i=1}^n x_i^2.$$