Probability excersice 

If $Z$ is a Gaussian random variable with mean $\mu_Z = 0$ and variance $\sigma^2_Z = 1$, and $Y$ is defined as: $$Y=a + bZ +cZ^2$$ for some constants $a, b, c$ show that the correlation coefficient of $Y$ and $Z$ is given by: $$\rho_{YZ}=\frac{b}{\sqrt{b^2+2c^2}}$$ 

I'm late. I have no idea to start and this is for tomorrow. I was on training and have no break to do this work. I only need to start
 A: You can check that $\Bbb{E}(Z^2)=1$, $\Bbb{E}(Z^3)=0$ and $\Bbb{E}(Z^4)=3$ by direct calculation or moment-generating function. So $\Bbb{E}(Y)=a+c$ and 
$$\Bbb{V}(Y)=\Bbb{E}(Y^2)-\Bbb{E}(Y)^2=(a^2+b^2+3c^2+2ac)-(a^2+c^2+2ac)=b^2+2c^2.$$
Therefore
$$
\rho_{YZ}=\frac{\Bbb{E}[(Y-\Bbb{E}(Y))(Z-\Bbb{E}(Z))]}{\sigma(Y)\sigma(Z)}=\frac{b}{\sqrt{b^2+2c^2}}.
$$
(Details of the proof are left to yours.)
A: A start: We need the covariance of $Y$ and $Z$. And we will need to divide that by $\sqrt{\text{Var}(Y)\text{Var}(Z)}$.
For the covariance, we need $E(YZ)-E(Y)E(Z)$, but luckily $E(Z)=0$.
And we know that $YZ=aZ+bZ^2+cZ^3$. The expectation of this is not bad, since two of the parts are very easy.
Now the bad part, we want the variance of $Y$, that is, of $a+bZ+cZ^2$. The $a$ doesn't matter. The variance will not be hard if we can find $E(bZ+cZ^2)^2$. Almost everything is easy, but we will need $E(Z^4)$. One can look this up, use the moment generating function if that has been computed, or set up the integral and calculate (integration by parts). 
A: Recall that
$$Cov(Y,A+B)=Cov(Y,A)+Cov(Y,B)$$,
that 
$$Cov(Y,c)=0$$ 
if $c$ is a constant, and that
$$Cov(Z,Z)=VZ.$$
