Solve 2x^2+nx+16=0

$$2 x ^ { 2 } + n x + 16 = 0$$

Solve for n

$$n=-2x-\frac{16}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$nx+16=-2x^{2}$$

Subtract 16 from both sides.

$$nx=-2x^{2}-16$$

The equation is in standard form.

$$xn=-2x^{2}-16$$

Divide both sides by x.

$$\frac{xn}{x}=\frac{-2x^{2}-16}{x}$$

Dividing by x undoes the multiplication by x.

$$n=\frac{-2x^{2}-16}{x}$$

Divide -2x^{2}-16 by x.

$$n=-2x-\frac{16}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{n^{2}-128}-n}{4}$$

$$x=\frac{-\sqrt{n^{2}-128}-n}{4}$$

Solve for x

$$x=\frac{\sqrt{n^{2}-128}-n}{4}$$

$$x=\frac{-\sqrt{n^{2}-128}-n}{4}\text{, }|n|\geq 8\sqrt{2}$$