Solve x^2-2(k+1)x-1-k=0

$$x ^ { 2 } - 2 ( k + 1 ) x - 1 - k = 0$$

Solve for k

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

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

Steps for Solving Linear Equation

Add 1 to both sides. Anything plus zero gives itself.

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

Use the distributive property to multiply -2 by k+1.

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

Use the distributive property to multiply -2k-2 by x.

$$x^{2}-2kx-2x-k=1$$

Subtract x^{2} from both sides.

$$-2kx-2x-k=1-x^{2}$$

Add 2x to both sides.

$$-2kx-k=1-x^{2}+2x$$

Combine all terms containing k.

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

The equation is in standard form.

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

Divide both sides by -2x-1.

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

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

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

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

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

Solve for x (complex solution)

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

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

Solve for x

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

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