Solve for k 4x^2-kx+4=0

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

Move all terms not containing $k$ to the right side of the equation.

Subtract $4 x^{2}$ from both sides of the equation.

$- k x + 4 = - 4 x^{2}$

Subtract $4$ from both sides of the equation.

$- k x = - 4 x^{2} - 4$

Divide each term in $- k x = - 4 x^{2} - 4$ by $- 1 x$ and simplify.

Divide each term in $- k x = - 4 x^{2} - 4$ by $- 1 x$ .

$\frac{- k x}{- 1 x} = \frac{- 4 x^{2}}{- 1 x} + \frac{- 4}{- 1 x}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{k x}{x} = \frac{- 4 x^{2}}{- 1 x} + \frac{- 4}{- 1 x}$

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{k x}{x} = \frac{- 4 x^{2}}{- 1 x} + \frac{- 4}{- 1 x}$

Divide $k$ by $1$ .

$k = \frac{- 4 x^{2}}{- 1 x} + \frac{- 4}{- 1 x}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $x^{2}$ and $x$ .

Factor $x$ out of $- 4 x^{2}$ .

$k = \frac{x (- 4 x)}{- 1 x} + \frac{- 4}{- 1 x}$

Move the negative one from the denominator of $\frac{- 4 x}{- 1}$ .

$k = - 1 \cdot (- 4 x) + \frac{- 4}{- 1 x}$

Rewrite $- 1 \cdot (- 4 x)$ as $- (- 4 x)$ .

$k = - (- 4 x) + \frac{- 4}{- 1 x}$

Multiply $- 4$ by $- 1$ .

$k = 4 x + \frac{- 4}{- 1 x}$

Dividing two negative values results in a positive value.

$k = 4 x + \frac{4}{x}$