$${ x }^{ 2 } +y(4x+y-1)+2=0$$
$x=\sqrt{\left(3y-2\right)\left(y+1\right)}-2y$
$x=-\sqrt{\left(3y-2\right)\left(y+1\right)}-2y$
$y=\frac{\sqrt{12x^{2}-8x-7}}{2}-2x+\frac{1}{2}$
$y=-\frac{\sqrt{12x^{2}-8x-7}}{2}-2x+\frac{1}{2}$
$x=\sqrt{\left(3y-2\right)\left(y+1\right)}-2y$
$x=-\sqrt{\left(3y-2\right)\left(y+1\right)}-2y\text{, }y\leq -1\text{ or }y\geq \frac{2}{3}$
$y=\frac{\sqrt{12x^{2}-8x-7}}{2}-2x+\frac{1}{2}$
$y=-\frac{\sqrt{12x^{2}-8x-7}}{2}-2x+\frac{1}{2}\text{, }x\leq -\frac{1}{2}\text{ or }x\geq \frac{7}{6}$