I have this theorem in my notes stating that $var(x)=E(x^2)-(MuX)^2$
the example with it is, given $MuX=5$ and $Var(x)=0.5$ then $E(x^2)=25.5$.
But what does $E(x^2)$ even mean?
I read elsewhere that it is the 2nd moment around the mean which is variance but the example clearly says variance is $0.5$.
Can it be converted to a simple expected value?
Please use MathJax, you question is hard to read.
What you have in the first line is the $\mathit{definition}$ of variance, from with you can easily find $25.5 = 5^2 + 0.5$, $\mathbf{E}X^2$ is indeed the second moment, while $\mathbf{Var} X$ is the second $\textit{standard}$ moment (i.e. standardized around the mean.
EDIT: the original question was 'But what does E(x2) even mean?', which is very vague. What I can say is that variance of the random variable measures the discrepancy/diversion/dispersion from the mean,$ \ \mu$. The second moment, $\mathbf{E}X^2 \ \textit{arises} $ in the definition of variance, so computation of variance (normalized second moment) is impossible without the knowledge of the second moment.