Variance of Bernoulli Distribution/Proof 2

Proof
From Variance as Expectation of Square minus Square of Expectation:
 * $\var X = \expect {X^2} - \paren {\expect X}^2$

From Expectation of Function of Discrete Random Variable:
 * $\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \, \map \Pr {X = x}$

So:

Then: