Solve yx^2+(y+2)x+4=0

$$y { x }^{ 2 } +(y+2)x+4=0$$

Solve for y

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

$$x\neq -1\text{ and }x\neq 0$$

Steps for Solving Linear Equation

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

$$yx^{2}+yx+2x+4=0$$

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

$$yx^{2}+yx+4=-2x$$

Subtract 4 from both sides.

$$yx^{2}+yx=-2x-4$$

Combine all terms containing y.

$$\left(x^{2}+x\right)y=-2x-4$$

Divide both sides by x^{2}+x.

$$\frac{\left(x^{2}+x\right)y}{x^{2}+x}=\frac{-2x-4}{x^{2}+x}$$

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

$$y=\frac{-2x-4}{x^{2}+x}$$

Divide -2x-4 by x^{2}+x.

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{y^{2}-12y+4}-y-2}{2y}\text{; }x=-\frac{\sqrt{y^{2}-12y+4}+y+2}{2y}\text{, }&y\neq 0\\x=-2\text{, }&y=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{y^{2}-12y+4}-y-2}{2y}\text{; }x=-\frac{\sqrt{y^{2}-12y+4}+y+2}{2y}\text{, }&\left(y\neq 0\text{ and }y\leq 6-4\sqrt{2}\right)\text{ or }y\geq 4\sqrt{2}+6\\x=-2\text{, }&y=0\end{matrix}\right.$$