Solve 3x^2+(y+2)x+y+1=0

$$3 x ^ { 2 } + ( y + 2 ) x + y + 1 = 0$$

Solve for y

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

$$x\neq -1$$

Steps for Solving Linear Equation

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

$$3x^{2}+yx+2x+y+1=0$$

Subtract 3x^{2} from both sides. Anything subtracted from zero gives its negation.

$$yx+2x+y+1=-3x^{2}$$

Subtract 2x from both sides.

$$yx+y+1=-3x^{2}-2x$$

Subtract 1 from both sides.

$$yx+y=-3x^{2}-2x-1$$

Combine all terms containing y.

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

Divide both sides by x+1.

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

Dividing by x+1 undoes the multiplication by x+1.

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

Divide -3x^{2}-2x-1 by x+1.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{y^{2}-8y-8}}{6}-\frac{y}{6}-\frac{1}{3}$$

$$x=-\frac{\sqrt{y^{2}-8y-8}}{6}-\frac{y}{6}-\frac{1}{3}$$

Solve for x

$$x=\frac{\sqrt{y^{2}-8y-8}}{6}-\frac{y}{6}-\frac{1}{3}$$

$$x=-\frac{\sqrt{y^{2}-8y-8}}{6}-\frac{y}{6}-\frac{1}{3}\text{, }y\geq 2\sqrt{6}+4\text{ or }y\leq 4-2\sqrt{6}$$