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

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

Solve for x

$$x=-\frac{3y^{2}}{4}+1$$

Steps for Solving Linear Equation

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

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

Add 4 to both sides.

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

The equation is in standard form.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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

$$x=-\frac{3y^{2}}{4}+1$$

Solve for y (complex solution)

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

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

Solve for y

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

$$y=-\frac{2\sqrt{3-3x}}{3}\text{, }x\leq 1$$