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

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

Solve for k

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

$$x\neq 2$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

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

Add 2x to both sides.

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

Subtract 4 from both sides.

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

Combine all terms containing k.

$$\left(-x+2\right)k=-x^{2}+2x-4$$

The equation is in standard form.

$$\left(2-x\right)k=-x^{2}+2x-4$$

Divide both sides by -x+2.

$$\frac{\left(2-x\right)k}{2-x}=\frac{-x^{2}+2x-4}{2-x}$$

Dividing by -x+2 undoes the multiplication by -x+2.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(k-6\right)\left(k+2\right)}+k+2}{2}$$

$$x=\frac{-\sqrt{\left(k-6\right)\left(k+2\right)}+k+2}{2}$$

Solve for x

$$x=\frac{\sqrt{\left(k-6\right)\left(k+2\right)}+k+2}{2}$$

$$x=\frac{-\sqrt{\left(k-6\right)\left(k+2\right)}+k+2}{2}\text{, }k\leq -2\text{ or }k\geq 6$$