Solve (y-1)^2=2x+8

$$( y - 1 ) ^ { 2 } = 2 x + 8$$

Solve for x

$$x=\frac{\left(y-1\right)^{2}-8}{2}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$y^{2}-2y+1=2x+8$$

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

$$2x+8=y^{2}-2y+1$$

Subtract 8 from both sides.

$$2x=y^{2}-2y+1-8$$

Subtract 8 from 1 to get -7.

$$2x=y^{2}-2y-7$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{y^{2}-2y-7}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{y^{2}-2y-7}{2}$$

Divide y^{2}-2y-7 by 2.

$$x=\frac{y^{2}}{2}-y-\frac{7}{2}$$

Solve for y (complex solution)

$$y=-\sqrt{2\left(x+4\right)}+1$$

$$y=\sqrt{2\left(x+4\right)}+1$$

Solve for y

$$y=-\sqrt{2\left(x+4\right)}+1$$

$$y=\sqrt{2\left(x+4\right)}+1\text{, }x\geq -4$$