Why does $ \operatorname{Var}(X) = E[X^2] - (E[X])^2 $ $ \operatorname{Var}(X) = E[X^2] - (E[X])^2 $
I have seen and understand (mathematically) the proof for this. What I want to understand is: intuitively, why is this true? What does this formula tell us? From the formula, we see that if we subtract the square of expected value of x from the expected value of $ x^2 $, we get a measure of dispersion in the data (or in the case of standard deviation, the root of this value gets us a measure of dispersion in the data). 
So it seems that there is some linkage between the expected value of $ x^2 $ and $ x $. How do I make sense of this formula? For example, the formula
$$ \sigma^2 = \frac 1n \sum_{i = 1}^n (x_i - \bar{x})^2 $$
makes perfect intuitive sense. It simply gives us the average of squares of deviations from the mean. What does the other formula tell us?

Easy! Expand by the definition. Variance is the mean squared deviation, i.e., $V(X) = E((X-\mu)^2).$ Now:
$$ (X-\mu)^2 = X^2 - 2X \mu + \mu^2$$
and use the fact that $E(\cdot)$ is a linear function and that $\mu$ (the mean) is a constant.
The shortcut computes the same thing, but counts the difference in the mean of squares and the square of the mean.
