Solve y^2-2y-4x+9=0

$${ y }^{ 2 } -2y-4x+9=0$$

Solve for x

$$x=\frac{y^{2}-2y+9}{4}$$

Steps for Solving Linear Equation

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

$$-2y-4x+9=-y^{2}$$

Add 2y to both sides.

$$-4x+9=-y^{2}+2y$$

Subtract 9 from both sides.

$$-4x=-y^{2}+2y-9$$

Divide both sides by -4.

$$\frac{-4x}{-4}=\frac{-y^{2}+2y-9}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x=\frac{-y^{2}+2y-9}{-4}$$

Divide -y^{2}+2y-9 by -4.

$$x=\frac{y^{2}}{4}-\frac{y}{2}+\frac{9}{4}$$

Solve for y (complex solution)

$$y=2\sqrt{x-2}+1$$

$$y=-2\sqrt{x-2}+1$$

Solve for y

$$y=2\sqrt{x-2}+1$$

$$y=-2\sqrt{x-2}+1\text{, }x\geq 2$$