Solve 2x^2+kx+k=0

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

Solve for k

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

$$x\neq -1$$

Steps for Solving Linear Equation

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

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

Combine all terms containing k.

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

Divide both sides by x+1.

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

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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{k\left(k-8\right)}-k}{4}$$

$$x=\frac{-\sqrt{k\left(k-8\right)}-k}{4}$$

Solve for x

$$x=\frac{\sqrt{k\left(k-8\right)}-k}{4}$$

$$x=\frac{-\sqrt{k\left(k-8\right)}-k}{4}\text{, }k\geq 8\text{ or }k\leq 0$$