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.
 A: 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.

A: If you want to solve with respect to $x$ that write the equation as:
$$
x^2+x(2y+3)+y^2-3y-18=0
$$
that use the quadratic formula
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$$
with
$$
a=1 \qquad b=2y+3 \qquad c=y^2-3y-18
$$
If you want to solve with respect to $y$ do the same ordering the equation in $y$.
A: Hint
$$
 \color{red}{x^2}+\color{red}{2x}y+y^2+\color{red}{3x}-3y+18=\color{red}{\left(x+y+\frac{3}{2}\right)^2-3y-y^2-\frac{9}{4}}+y^2-3y-18
$$
Find a second square end solve the equation.
A: You can use the transformation
 x=1/√2(a+b)
y=1/√2(-a+b)
