Solve 3x^2-yx+4=0

$$3 { x }^{ 2 } -yx+4=0$$

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.

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

Subtract 4 from both sides.

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

The equation is in standard form.

$$\left(-x\right)y=-3x^{2}-4$$

Divide both sides by -x.

$$\frac{\left(-x\right)y}{-x}=\frac{-3x^{2}-4}{-x}$$

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

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

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