Solve x^2+2nx*x+3=0

$$x ^ { 2 } + 2 n x \cdot x + 3 = 0$$

Solve for n

$$n=-\frac{1}{2}-\frac{3}{2x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply x and x to get x^{2}.

$$x^{2}+2nx^{2}+3=0$$

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

$$2nx^{2}+3=-x^{2}$$

Subtract 3 from both sides.

$$2nx^{2}=-x^{2}-3$$

The equation is in standard form.

$$2x^{2}n=-x^{2}-3$$

Divide both sides by 2x^{2}.

$$\frac{2x^{2}n}{2x^{2}}=\frac{-x^{2}-3}{2x^{2}}$$

Dividing by 2x^{2} undoes the multiplication by 2x^{2}.

$$n=\frac{-x^{2}-3}{2x^{2}}$$

Divide -x^{2}-3 by 2x^{2}.

$$n=-\frac{1}{2}-\frac{3}{2x^{2}}$$

Solve for x (complex solution)

$$x=-\sqrt{3}i\left(2n+1\right)^{-\frac{1}{2}}$$

$$x=\sqrt{3}i\left(2n+1\right)^{-\frac{1}{2}}\text{, }n\neq -\frac{1}{2}$$

Solve for x

$$x=\sqrt{-\frac{3}{2n+1}}$$

$$x=-\sqrt{-\frac{3}{2n+1}}\text{, }n<-\frac{1}{2}$$