exercise - discrete random variables

Given $E(X^2)=\frac{5}{2}$, find the variance.

x	-1	0	1	2	3
Pr(X=x)	m	2m	n	0.5n	n+m
x^2	1	0	1	4	9

$4m+2.5n=1$
$\frac{5}{2}=m+n+2n+9n+9m$
$\frac{5}{2}=10m+12n$
$\frac{5}{2}=10m+6.25n$

$E(X)=(\frac{-1}{4}+\frac{3}{4}{)}^2=\frac{1}{4}$
$\text{Var}(X)=\frac{5}{2}-\frac{1}{4}=\frac{9}{4}$