Solve 3y^2-2x-3y-7=0

$$3 { y }^{ 2 } -2x-3y-7=0$$

Solve for x

$$x=\frac{3y^{2}-3y-7}{2}$$

Steps for Solving Linear Equation

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

$$-2x-3y-7=-3y^{2}$$

Add 3y to both sides.

$$-2x-7=-3y^{2}+3y$$

Add 7 to both sides.

$$-2x=-3y^{2}+3y+7$$

The equation is in standard form.

$$-2x=7+3y-3y^{2}$$

Divide both sides by -2.

$$\frac{-2x}{-2}=\frac{7+3y-3y^{2}}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x=\frac{7+3y-3y^{2}}{-2}$$

Divide -3y^{2}+3y+7 by -2.

$$x=\frac{3y^{2}-3y-7}{2}$$

Solve for y (complex solution)

$$y=\frac{\sqrt{24x+93}}{6}+\frac{1}{2}$$

$$y=-\frac{\sqrt{24x+93}}{6}+\frac{1}{2}$$

Solve for y

$$y=\frac{\sqrt{24x+93}}{6}+\frac{1}{2}$$

$$y=-\frac{\sqrt{24x+93}}{6}+\frac{1}{2}\text{, }x\geq -\frac{31}{8}$$