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

$$( y + 3 ) ^ { 2 } = - 8 ( x + 2 )$$

Solve for x

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

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$y^{2}+6y+9=-8\left(x+2\right)$$

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

$$y^{2}+6y+9=-8x-16$$

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

$$-8x-16=y^{2}+6y+9$$

Add 16 to both sides.

$$-8x=y^{2}+6y+9+16$$

Add 9 and 16 to get 25.

$$-8x=y^{2}+6y+25$$

Divide both sides by -8.

$$\frac{-8x}{-8}=\frac{y^{2}+6y+25}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x=\frac{y^{2}+6y+25}{-8}$$

Divide y^{2}+6y+25 by -8.

$$x=-\frac{y^{2}}{8}-\frac{3y}{4}-\frac{25}{8}$$

Solve for y (complex solution)

$$y=-2\sqrt{-2x-4}-3$$

$$y=2\sqrt{-2x-4}-3$$

Solve for y

$$y=-2\sqrt{-2x-4}-3$$

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