Solve 2x^2+mx+8=0

$$2 x ^ { 2 } + m x + 8 = 0$$

Solve for m

$$m=-2x-\frac{8}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$mx+8=-2x^{2}$$

Subtract 8 from both sides.

$$mx=-2x^{2}-8$$

The equation is in standard form.

$$xm=-2x^{2}-8$$

Divide both sides by x.

$$\frac{xm}{x}=\frac{-2x^{2}-8}{x}$$

Dividing by x undoes the multiplication by x.

$$m=\frac{-2x^{2}-8}{x}$$

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

$$m=-2x-\frac{8}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{m^{2}-64}-m}{4}$$

$$x=\frac{-\sqrt{m^{2}-64}-m}{4}$$

Solve for x

$$x=\frac{\sqrt{m^{2}-64}-m}{4}$$

$$x=\frac{-\sqrt{m^{2}-64}-m}{4}\text{, }|m|\geq 8$$