How can I solve $x^2+2xy+y^2+3x-3y-18=0$ $$x^2+2xy+y^2+3x-3y-18=0$$
I don't know how to solve it.

When a general conic equation
$$ Ax^2+Bxy+Cy^2+Dx +Ey +F=0$$
has $B\ne0$ and $A=C$ then rotating it $45^\circ$ will eliminate the $xy$ term.
This will be accomplished by the substitution
\begin{eqnarray}
x&=&\frac{X-Y}{\sqrt{2}}\\
y&=&\frac{(X+Y)}{\sqrt{2}}
\end{eqnarray}
\begin{equation}
x^2+2xy+y^2+3x-3y-18=0
\end{equation}
\begin{equation}
\frac{(X-Y)^2}{2}+(X-Y)(X+Y)+\frac{(X+Y)^2}{2}+\frac{3}{\sqrt{2}}(X-Y)-\frac{3}{\sqrt{2}}(X+Y)-18=0
\end{equation}
$$X^2=\frac{3}{\sqrt{2}}Y+18$$
So it is the equation of a parabola rotated $45^\circ$ about the origin.

