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

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

Solve for k

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

$$x\neq \frac{1}{2}$$

Steps for Solving Linear Equation

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

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

Subtract 2 from both sides.

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

Combine all terms containing k.

$$\left(-2x+1\right)k=-x^{2}-2$$

The equation is in standard form.

$$\left(1-2x\right)k=-x^{2}-2$$

Divide both sides by -2x+1.

$$\frac{\left(1-2x\right)k}{1-2x}=\frac{-x^{2}-2}{1-2x}$$

Dividing by -2x+1 undoes the multiplication by -2x+1.

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

Divide -x^{2}-2 by -2x+1.

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

Solve for x (complex solution)

$$x=\sqrt{\left(k-2\right)\left(k+1\right)}+k$$

$$x=-\sqrt{\left(k-2\right)\left(k+1\right)}+k$$

Solve for x

$$x=\sqrt{\left(k-2\right)\left(k+1\right)}+k$$

$$x=-\sqrt{\left(k-2\right)\left(k+1\right)}+k\text{, }k\leq -1\text{ or }k\geq 2$$