Question

$$y= { x }^{ 2 } +4x$$

Solve for x (complex solution)

$x=-\sqrt{y+4}-2$
$x=\sqrt{y+4}-2$

Solve for x

$x=-\sqrt{y+4}-2$
$x=\sqrt{y+4}-2\text{, }y\geq -4$

Solve for y

$y=x\left(x+4\right)$