Suppose $E[X] = 3$, $Var(X) = 1$, $E[Y] = 2$, $Var(Y) = \frac14$, $Corr(X,Y) = \frac12$
What is $E((X+2Y)^2)$?
We can rewrite $E((X+2Y)^2)$ as $E(X^2+4XY+4Y^2)$
We can solve for $E(X^2)$ using $Var(X) = E[X^2] - E[X]^2$. The same can be done for $E(Y^2)$
This gives us $E[X^2] = 10$ and $E[Y^2] = \frac{17}4$
We can compute $E[XY]$ through the expectation formula of covariance.
$Corr(X,Y) = \frac{Cov(X,Y)}{\sigma_x\sigma_y} \rightarrow \frac12 = \frac{Cov(X,Y)}{1\cdot\frac12} \rightarrow Cov(X,Y) = \frac14$
$Cov(X,Y) = E[X,Y] - E[X]E[Y] \rightarrow \frac14 = E[X,Y] - 3 \cdot 2 \rightarrow E[X,Y] = \frac{25}4$
$E[X^2] + E[4XY] + E[4Y^2] = 10 + 25 + 17 = 52$
Do you see any problems with this computation?
Here is how I compute $E[XY]$. $$ \rho = \dfrac{ cov(X,Y) }{ \sigma_x \sigma_y } $$ $$ \sigma_x = 1 $$ $$ \sigma_y = \dfrac{1}{2} $$ $$ \dfrac{1}{2} = \dfrac{ cov(X,Y) }{ \dfrac{1}{2} } $$ $$ \dfrac{1}{2} = \dfrac{ cov(X,Y) }{ \dfrac{1}{2} } $$ $$ cov(X,Y) = \dfrac{1}{4} $$ $$ cov(X,Y) = E(XY) - E(X)E(Y) $$ $$ \dfrac{1}{4} = E[XY] - 3(2) $$ $$ E[XY] = \dfrac{25}{4} $$
I now agree with your answer.