Solve (x^2+8)+(x+2)j=0

$$( x ^ { 2 } + 8 ) + ( x + 2 ) j = 0$$

Solve for j

$$j=-\frac{x^{2}+8}{x+2}$$

$$x\neq -2$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+2 by j.

$$x^{2}+8+xj+2j=0$$

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

$$8+xj+2j=-x^{2}$$

Subtract 8 from both sides.

$$xj+2j=-x^{2}-8$$

Combine all terms containing j.

$$\left(x+2\right)j=-x^{2}-8$$

Divide both sides by x+2.

$$\frac{\left(x+2\right)j}{x+2}=\frac{-x^{2}-8}{x+2}$$

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

$$j=\frac{-x^{2}-8}{x+2}$$

Divide -x^{2}-8 by x+2.

$$j=-\frac{x^{2}+8}{x+2}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{j^{2}-8j-32}-j}{2}$$

$$x=\frac{-\sqrt{j^{2}-8j-32}-j}{2}$$

Solve for x

$$x=\frac{\sqrt{j^{2}-8j-32}-j}{2}$$

$$x=\frac{-\sqrt{j^{2}-8j-32}-j}{2}\text{, }j\geq 4\sqrt{3}+4\text{ or }j\leq 4-4\sqrt{3}$$