Obtain variance from density of random variable X is a random variable with the density
$$F_X(x)  = \begin{cases}2 x^{-2}, & x \in (1,2)\\
0 & \operatorname{otherwise}\end{cases}$$
How can I find the 
$VAR(3X^2-5)$
Normal variation is computed from equation 
$E(X^2)-(EX)^2 = VAR^2(X)$
So instead of multiplication my function by X i need to multiplicate by $3X^2-5$?
$E(3X^2-5)^2 =\int_1^2(3x^2-5)^2*2x^{-2} =15.5
$
$(EX)^2 =\int_1^2(3x^2-5)*2x^{-2} =1$
$VAR(3X^2-5)=\sqrt{14.5}$
Should it be like that?

First, using the fact that $Var(aX)=a^2Var(X)$ and $Var(X+c)=Var(X)$, note that $Var(3X^2-5)=9Var(X^2)$ and $Var(X^2)=E(X^4)-(E(X^2))^2$. 
$E(X^4)=\int_1^2 x^4 .2x^{-2}=14/3$, similarly $E(X^2)=2$ and so $Var(X^2)=2/3$ 
and $Var(3X^2-5)=6$
Your approach is also correct but you are making a calculation error, that is $Var(3X^2-5)=E((3X^2-5)^2) - (E(3X^2-5))^2$ where $E((3X^2-5)^2)=\int_1^2(3x^2-5)2x^{-2}=7$ and $E(3X^2-5)=1$ and so $Var(3X^2-5)=6$.  .
