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

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

Solve for y

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

$$x\neq 0$$

Steps for Solving Linear Equation

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

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

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

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

Subtract 3 from both sides.

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

Divide both sides by 2x^{2}.

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

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

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

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

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

Solve for x (complex solution)

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

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

Solve for x

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

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