Solve left(y-4right)^2=8(x+3)

$${ \left(y-4 \right) }^{ 2 } =8(x+3)$$

Solve for x

$$x=\frac{\left(y-4\right)^{2}-24}{8}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$y^{2}-8y+16=8\left(x+3\right)$$

Use the distributive property to multiply 8 by x+3.

$$y^{2}-8y+16=8x+24$$

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

$$8x+24=y^{2}-8y+16$$

Subtract 24 from both sides.

$$8x=y^{2}-8y+16-24$$

Subtract 24 from 16 to get -8.

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

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

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

$$x=\frac{y^{2}}{8}-y-1$$

Solve for y (complex solution)

$$y=-2\sqrt{2x+6}+4$$

$$y=2\sqrt{2x+6}+4$$

Solve for y

$$y=-2\sqrt{2x+6}+4$$

$$y=2\sqrt{2x+6}+4\text{, }x\geq -3$$