Solve 3x^2+xy+4=0

$$3 x ^ { 2 } + x y + 4 = 0$$

Solve for y

$$y=-3x-\frac{4}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$xy+4=-3x^{2}$$

Subtract 4 from both sides.

$$xy=-3x^{2}-4$$

Divide both sides by x.

$$\frac{xy}{x}=\frac{-3x^{2}-4}{x}$$

Dividing by x undoes the multiplication by x.

$$y=\frac{-3x^{2}-4}{x}$$

Divide -3x^{2}-4 by x.

$$y=-3x-\frac{4}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y^{2}-48}-y}{6}$$

$$x=\frac{-\sqrt{y^{2}-48}-y}{6}$$

Solve for x

$$x=\frac{\sqrt{y^{2}-48}-y}{6}$$

$$x=\frac{-\sqrt{y^{2}-48}-y}{6}\text{, }|y|\geq 4\sqrt{3}$$