Solve 6x-3y^2-12y+4=0

$$6 x - 3 y ^ { 2 } - 12 y + 4 = 0$$

Solve for x

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

Steps for Solving Linear Equation

Add 3y^{2} to both sides. Anything plus zero gives itself.

$$6x-12y+4=3y^{2}$$

Add 12y to both sides.

$$6x+4=3y^{2}+12y$$

Subtract 4 from both sides.

$$6x=3y^{2}+12y-4$$

Divide both sides by 6.

$$\frac{6x}{6}=\frac{3y^{2}+12y-4}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x=\frac{3y^{2}+12y-4}{6}$$

Divide 3y^{2}+12y-4 by 6.

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

Solve for y (complex solution)

$$y=-\frac{\sqrt{18x+48}}{3}-2$$

$$y=\frac{\sqrt{18x+48}}{3}-2$$

Solve for y

$$y=-\frac{\sqrt{18x+48}}{3}-2$$

$$y=\frac{\sqrt{18x+48}}{3}-2\text{, }x\geq -\frac{8}{3}$$