Solve (y-2)^2=2x-8-4

$$( y - 2 ) ^ { 2 } = 2 x - 8 - 4$$

Solve for x

$$x=\frac{\left(y-2\right)^{2}+12}{2}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-2\right)^{2}.

$$y^{2}-4y+4=2x-8-4$$

Subtract 4 from -8 to get -12.

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

Swap sides so that all variable terms are on the left hand side.

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

Add 12 to both sides.

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

Add 4 and 12 to get 16.

$$2x=y^{2}-4y+16$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{y^{2}-4y+16}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{y^{2}-4y+16}{2}$$

Divide y^{2}-4y+16 by 2.

$$x=\frac{y^{2}}{2}-2y+8$$

Solve for y (complex solution)

$$y=-\sqrt{2x-12}+2$$

$$y=\sqrt{2x-12}+2$$

Solve for y

$$y=-\sqrt{2x-12}+2$$

$$y=\sqrt{2x-12}+2\text{, }x\geq 6$$