Why is $E[(X − b) ^2 ]$ minimal when $b = µ$?

Question

If $X$ is a random variable with a mean µ and a variance $σ^2$, why is $$E[(X − b) ^2 ]$$ minimal when $b = µ$?

Answer 1

$$ \mathbb E[(X-b)^2] = \mathbb E[X^2 - 2 b X + b^2] = \mathbb E[X^2] - 2 b \mathbb E[X] + b^2 = \sigma^2 + (b-\mu)^2$$