Solve for k x^2-kx+169=0

$x^{2} - k x + 169 = 0$

Step 1

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

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

$- k x + 169 = - x^{2}$

Subtract $169$ from both sides of the equation.

$- k x = - x^{2} - 169$

Step 2

Divide each term in $- k x = - x^{2} - 169$ by $- 1 x$ and simplify.

Divide each term in $- k x = - x^{2} - 169$ by $- 1 x$ .

$\frac{- k x}{- 1 x} = \frac{- x^{2}}{- 1 x} + \frac{- 169}{- 1 x}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{k x}{x} = \frac{- x^{2}}{- 1 x} + \frac{- 169}{- 1 x}$

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{k x}{x} = \frac{- x^{2}}{- 1 x} + \frac{- 169}{- 1 x}$

Divide $k$ by $1$ .

$k = \frac{- x^{2}}{- 1 x} + \frac{- 169}{- 1 x}$

Simplify the right side.

Simplify each term.

Dividing two negative values results in a positive value.

$k = \frac{x^{2}}{x} + \frac{- 169}{- 1 x}$

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

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

$k = \frac{x \cdot x}{x} + \frac{- 169}{- 1 x}$

Cancel the common factors.

Raise $x$ to the power of $1$ .

$k = \frac{x \cdot x}{x^{1}} + \frac{- 169}{- 1 x}$

Factor $x$ out of $x^{1}$ .

$k = \frac{x \cdot x}{x \cdot 1} + \frac{- 169}{- 1 x}$

Cancel the common factor.

$k = \frac{x \cdot x}{x \cdot 1} + \frac{- 169}{- 1 x}$

Rewrite the expression.

$k = \frac{x}{1} + \frac{- 169}{- 1 x}$

Divide $x$ by $1$ .

$k = x + \frac{- 169}{- 1 x}$

Dividing two negative values results in a positive value.

$k = x + \frac{169}{x}$