Solving the nonlinear Diophantine equation $x^2-3x=2y^2$ How can I solve (find all the solutions) the nonlinear Diophantine equation $x^2-3x=2y^2$? 
I included here what I had done so far. Thanks for your help.
Note: The equation above can be rewritten into $x^2-3x-2y^2=0$ which is quadratic in $x$. By quadratic formula we have the following solutions for $x$.
\begin{equation}
x=\frac{3\pm\sqrt{9+8y^2}}{2}
\end{equation}
I want $x$ to be a positive integer so I will just consider:
\begin{equation}
x=\frac{3+\sqrt{9+8y^2}}{2}
\end{equation}   
From here I don't know how to proceed but after trying out values for $1\leq y\leq 1000$ I only have the following $y$ that yields a positive integer $x$.
\begin{equation}
y=\{0,3,18,105,612\}.
\end{equation}
Again, thanks for any help.   

It is evident that equation.
$$x^2-3x=2y^2$$
This is a Pell equation. And we have to use solutions which are determined by the sequence. The next value from the previous one.
$$p_2=3p_1+4s_1$$
$$s_2=2p_1+3s_1$$
If we use the sequence where the first number. $(p_1 ;s_1) - (1 ; 1)$
Decisions will be.
$$x=6s^2$$
$$y=3ps$$
If we use the sequence where the first number.  $(p_1 ;s_1) - (3 ; 2)$
Decisions will be.
$$x=3p^2$$
$$y=3ps$$
