Question

$$4x^{2}-(k-2)x+9=0$$

Solve for k

$k=4x+2+\frac{9}{x}$
$x\neq 0$

Show Solution

Solve for x (complex solution)

$x=\frac{\sqrt{\left(k-14\right)\left(k+10\right)}+k-2}{8}$
$x=\frac{-\sqrt{\left(k-14\right)\left(k+10\right)}+k-2}{8}$

Solve for x

$x=\frac{\sqrt{\left(k-14\right)\left(k+10\right)}+k-2}{8}$
$x=\frac{-\sqrt{\left(k-14\right)\left(k+10\right)}+k-2}{8}\text{, }k\leq -10\text{ or }k\geq 14$