Solve for k x^2-3kx+3k-7=0

$x^{2} - 3 k x + 3 k - 7 = 0$

Step 1

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

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

$- 3 k x + 3 k - 7 = - x^{2}$

Add $7$ to both sides of the equation.

$- 3 k x + 3 k = - x^{2} + 7$

Step 2

Factor $3 k$ out of $- 3 k x + 3 k$ .

Factor $3 k$ out of $- 3 k x$ .

$3 k (- 1 x) + 3 k = - x^{2} + 7$

Factor $3 k$ out of $3 k$ .

$3 k (- 1 x) + 3 k (1) = - x^{2} + 7$

Factor $3 k$ out of $3 k (- 1 x) + 3 k (1)$ .

$3 k (- 1 x + 1) = - x^{2} + 7$

Step 3

Rewrite $- 1 x$ as $- x$ .

$3 k (- x + 1) = - x^{2} + 7$

Step 4

Divide each term in $3 k (- x + 1) = - x^{2} + 7$ by $3 (- x + 1)$ and simplify.

Divide each term in $3 k (- x + 1) = - x^{2} + 7$ by $3 (- x + 1)$ .

$\frac{3 k (- x + 1)}{3 (- x + 1)} = \frac{- x^{2}}{3 (- x + 1)} + \frac{7}{3 (- x + 1)}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 k (- x + 1)}{3 (- x + 1)} = \frac{- x^{2}}{3 (- x + 1)} + \frac{7}{3 (- x + 1)}$

Rewrite the expression.

$\frac{k (- x + 1)}{- x + 1} = \frac{- x^{2}}{3 (- x + 1)} + \frac{7}{3 (- x + 1)}$

Cancel the common factor of $- x + 1$ .

Cancel the common factor.

$\frac{k (- x + 1)}{- x + 1} = \frac{- x^{2}}{3 (- x + 1)} + \frac{7}{3 (- x + 1)}$

Divide $k$ by $1$ .

$k = \frac{- x^{2}}{3 (- x + 1)} + \frac{7}{3 (- x + 1)}$

Simplify the right side.

Move the negative in front of the fraction.

$k = - \frac{x^{2}}{3 (- x + 1)} + \frac{7}{3 (- x + 1)}$