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

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

Solve for x

$$x=\frac{\left(y-5\right)\left(y+3\right)}{8}$$

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}.

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

Use the distributive property to multiply 8 by x+2.

$$y^{2}-2y+1=8x+16$$

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

$$8x+16=y^{2}-2y+1$$

Subtract 16 from both sides.

$$8x=y^{2}-2y+1-16$$

Subtract 16 from 1 to get -15.

$$8x=y^{2}-2y-15$$

Divide both sides by 8.

$$\frac{8x}{8}=\frac{\left(y-5\right)\left(y+3\right)}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x=\frac{\left(y-5\right)\left(y+3\right)}{8}$$

Solve for y (complex solution)

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

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

Solve for y

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

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