Prove that $x^2 + 4xy + 4y^2 + 3x + 6y - 2 = 0$ is the equation of two straight lines 
Prove that the equation $x^2 + 4xy + 4y^2 + 3x + 6y - 2 = 0$ is of two straight lines. Find each line's equation, and find the angle between the two lines.

My Attempt:
Factorize $4y^2 + 6y - 2$; Can't be factorized without quadratic formula.
Thanks in advance.

The given equation can be written as 
$$ (x+2y)^2+3(x+2y)-2 = 0 $$
or as
$$ (x+2y)(x+2y+3) = 2 $$
and the only solutions of $z(z+3)=2$ are given by $z=\frac{-3\pm\sqrt{17}}{2}$, so our locus is given by the union of the parallel lines $x+2y=\frac{-3+\sqrt{17}}{2}$ and $x+2y=\frac{-3-\sqrt{17}}{2}$.
