Solve (y-3)^2=8x-8

$$( y - 3 ) ^ { 2 } = 8 x - 8$$

Solve for x

$$x=\frac{\left(y-3\right)^{2}+8}{8}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$y^{2}-6y+9=8x-8$$

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

$$8x-8=y^{2}-6y+9$$

Add 8 to both sides.

$$8x=y^{2}-6y+9+8$$

Add 9 and 8 to get 17.

$$8x=y^{2}-6y+17$$

Divide both sides by 8.

$$\frac{8x}{8}=\frac{y^{2}-6y+17}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x=\frac{y^{2}-6y+17}{8}$$

Divide y^{2}-6y+17 by 8.

$$x=\frac{y^{2}}{8}-\frac{3y}{4}+\frac{17}{8}$$

Solve for y (complex solution)

$$y=-2\sqrt{2\left(x-1\right)}+3$$

$$y=2\sqrt{2\left(x-1\right)}+3$$

Solve for y

$$y=-2\sqrt{2\left(x-1\right)}+3$$

$$y=2\sqrt{2\left(x-1\right)}+3\text{, }x\geq 1$$