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

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

Solve for y

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

Steps for Solving Linear Equation

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

$$x^{2}+4x+4=-8\left(y-1\right)$$

Use the distributive property to multiply -8 by y-1.

$$x^{2}+4x+4=-8y+8$$

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

$$-8y+8=x^{2}+4x+4$$

Subtract 8 from both sides.

$$-8y=x^{2}+4x+4-8$$

Subtract 8 from 4 to get -4.

$$-8y=x^{2}+4x-4$$

Divide both sides by -8.

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

Dividing by -8 undoes the multiplication by -8.

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

Divide x^{2}+4x-4 by -8.

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

Solve for x (complex solution)

$$x=-2\sqrt{2-2y}-2$$

$$x=2\sqrt{2-2y}-2$$

Solve for x

$$x=-2\sqrt{2-2y}-2$$

$$x=2\sqrt{2-2y}-2\text{, }y\leq 1$$