Solve left(y-1right)^2=8left(x+5right)

$${ \left(y-1 \right) }^{ 2 } =8 \left( x+5 \right)$$

Solve for x

$$x=\frac{\left(y-1\right)^{2}-40}{8}$$

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}.

$$y^{2}-2y+1=8\left(x+5\right)$$

Use the distributive property to multiply 8 by x+5.

$$y^{2}-2y+1=8x+40$$

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

$$8x+40=y^{2}-2y+1$$

Subtract 40 from both sides.

$$8x=y^{2}-2y+1-40$$

Subtract 40 from 1 to get -39.

$$8x=y^{2}-2y-39$$

Divide both sides by 8.

$$\frac{8x}{8}=\frac{y^{2}-2y-39}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x=\frac{y^{2}-2y-39}{8}$$

Divide y^{2}-2y-39 by 8.

$$x=\frac{y^{2}}{8}-\frac{y}{4}-\frac{39}{8}$$

Solve for y (complex solution)

$$y=-2\sqrt{2\left(x+5\right)}+1$$

$$y=2\sqrt{2\left(x+5\right)}+1$$

Solve for y

$$y=-2\sqrt{2\left(x+5\right)}+1$$

$$y=2\sqrt{2\left(x+5\right)}+1\text{, }x\geq -5$$