I have an equation that looks like this:
$11.16^2 = X^2 + Y^2 \implies 124.5456 - X^2 = Y^2 \implies 124.5456 - E(X^2) = E(Y^2)$ is that correct?
The X is random variable that is distributed by continuous uniform distribution from 1.43 to 9.46 . Y is also random variable.
So how could I get $E(X^2)$ from $E(X)$ to find $E(Y^2)$?
If you know the variance of $X$ then you can use the equation,$$ Var(X)=E[X^2]-(E[X])^2$$ to get the value of $E(X^2)$.
But, it's not necessary that you have to get $E(X^2)$ from $E(X)$ only. In general for any random variable $X$ with probability density function $f(x)$ and any measurable function $g(X)$ of $X$, expectation/expected value of $g(X)$ is $$ E[g(X)]=\int_{-\infty}^{\infty} \! g(x)f(x) \, \mathrm{d}x.$$
substituting $g(X)=X^2$ you can get the value of $E(X^2)$. In your case $f(x)=\frac{1}{9.46-1.43}$ for $1.43 \leq x\leq 9.46$ and $f(x)=0$ for other values of $x$ and thus integration domain is from 1.43 to 9.46.