Solve $x^2+xy+y^2+ \sqrt{3}y + 1=0$ for $xy$. I have a problem on my textbook:

$x,y \in\mathbb{R}$ and  $x^2+xy+y^2+ \sqrt{3}y + 1=0$ is given. Find the value of $xy$. 

I couldn't find.

Setting $t=xy\in\mathbb R$, the equation becomes (knowing that $y\ne0$):
$$\frac{t^2}{y^2}+t+y^2+\sqrt3y+1=0.$$
The discriminant of this quadratic equation is $$\Delta=b^2-4ac=\frac{-3y^2-4\sqrt3y-4}{y^2}=-3\left(1-\dfrac2{\sqrt3y}\right)^2$$ and for a real solution to exist, it must be zero.
Then 
$$xy=t=-\frac b{2a}=-\frac{y^2}2=-\frac23.$$
