Question

$$x^{2}+4x+l=0$$

Solve for l

$l=-x\left(x+4\right)$

Show Solution

Solve for x (complex solution)

$x=\sqrt{4-l}-2$
$x=-\sqrt{4-l}-2$

Solve for x

$x=\sqrt{4-l}-2$
$x=-\sqrt{4-l}-2\text{, }l\leq 4$