Solve -(x+2)^2=kx+2k+2

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

Solve for k

$$k=-\frac{x^{2}+4x+6}{x+2}$$

$$x\neq -2$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

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

To find the opposite of x^{2}+4x+4, find the opposite of each term.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2 from both sides.

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

Subtract 2 from -4 to get -6.

$$kx+2k=-x^{2}-4x-6$$

Combine all terms containing k.

$$\left(x+2\right)k=-x^{2}-4x-6$$

Divide both sides by x+2.

$$\frac{\left(x+2\right)k}{x+2}=\frac{-x^{2}-4x-6}{x+2}$$

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

$$k=\frac{-x^{2}-4x-6}{x+2}$$

Divide -x^{2}-4x-6 by x+2.

$$k=-\frac{x^{2}+4x+6}{x+2}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{k^{2}-8}}{2}-\frac{k}{2}-2$$

$$x=\frac{\sqrt{k^{2}-8}}{2}-\frac{k}{2}-2$$

Solve for x

$$x=-\frac{\sqrt{k^{2}-8}}{2}-\frac{k}{2}-2$$

$$x=\frac{\sqrt{k^{2}-8}}{2}-\frac{k}{2}-2\text{, }|k|\geq 2\sqrt{2}$$