Solve y^2-3x-4=0

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

Solve for x

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

Steps for Solving Linear Equation

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

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

Add 4 to both sides.

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

The equation is in standard form.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

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

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

Solve for y (complex solution)

$$y=-\sqrt{3x+4}$$

$$y=\sqrt{3x+4}$$

Solve for y

$$y=\sqrt{3x+4}$$

$$y=-\sqrt{3x+4}\text{, }x\geq -\frac{4}{3}$$