Question

$$3y^{2}+ky+12=0$$

Solve for k

$k=-3y-\frac{12}{y}$
$y\neq 0$

Show Solution

Solve for y

$y=\frac{\sqrt{k^{2}-144}-k}{6}$
$y=\frac{-\sqrt{k^{2}-144}-k}{6}\text{, }|k|\geq 12$