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

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

Solve for x

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

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.

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

Add 24 to both sides.

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

Add 16 and 24 to get 40.

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

Divide both sides by -8.

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

Dividing by -8 undoes the multiplication by -8.

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

Divide y^{2}-8y+40 by -8.

$$x=-\frac{y^{2}}{8}+y-5$$

Solve for y (complex solution)

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

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

Solve for y

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

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