Question: Per Wikipedia, this is what happens when you expand the variance equation:

$\ Var(X)=E[X-E[X])^2 $
$ =E[X^2 -2XE[X] + (E[X]))^2 $
$ =E[X^2] - 2E[X]E[X] + (E[X]))^2 $

How do we get the $\ 2E[X]E[X] $ term?

Answer: 
$$\begin{align}
\text{Var}(X)&=E\Big[\big(X-E[X]\big)^{2}\Big]\\
&=E\Big[X^{2}-2XE[X]+\big(E[X]\big)^{2}\Big]\\
&=E\big[X^{2}\big]-2E\big[XE[X]\big]+E\Big[\big(E[X]\big)^{2}\Big]\\
&\overset{\*}{=}E\big[X^{2}\big]-2E\big[XE[X]\big]+\big(E[X]\big)^{2}\\
&\overset{\*\*}{=}E\big[X^{2}\big]-2E[X]E[X]+\big(E[X]\big)^{2}\\
&=E\big[X^{2}\big]-2\big(E[X]\big)^{2}+\big(E[X]\big)^{2}\\
&=E\big[X^{2}\big]-\big(E[X]\big)^{2}
\end{align}$$

$^{\*}$Because $\big(E[X]\big)^{2}$ is a constant, so taking the expectation of $\big(E[X]\big)^{2}$ just gives the constant.

$^{\*\*}$Because $E[X]$ is a constant, so taking the expectation of $E[X]$ just gives the constant.
It's from linearity of expectation and the fact that $E[X]$ is constant. I'll let $E[X]=c$ at some steps for clarity.
\begin{align\*}
Var(X) &= E[ (X- E[X])^2] \\
&= E[ X^2 - 2X E[X] + (E[X])^2] \\
&= E[X^2] - E[2X E[X]] + E[(E[X])^2] \\
&= E[X^2] - E[2X c] + E[c^2] \\
&= E[X^2] - 2c E[X] + c^2 \\
&= E[X^2] - 2(E[X])^2 + (E[X])^2 \\
&= E[X^2] - (E[X])^2.
\end{align\*}