Solve (4+y)^2+2y(4+y)-y^2-z=0

$$( 4 + y ) ^ { 2 } + 2 y ( 4 + y ) - y ^ { 2 } - z = 0$$

Solve for z

$$z=2\left(y^{2}+8y+8\right)$$

Steps for Solving Linear Equation

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

$$16+8y+y^{2}+2y\left(4+y\right)-y^{2}-z=0$$

Use the distributive property to multiply 2y by 4+y.

$$16+8y+y^{2}+8y+2y^{2}-y^{2}-z=0$$

Combine 8y and 8y to get 16y.

$$16+16y+y^{2}+2y^{2}-y^{2}-z=0$$

Combine y^{2} and 2y^{2} to get 3y^{2}.

$$16+16y+3y^{2}-y^{2}-z=0$$

Combine 3y^{2} and -y^{2} to get 2y^{2}.

$$16+16y+2y^{2}-z=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$16y+2y^{2}-z=-16$$

Subtract 16y from both sides.

$$2y^{2}-z=-16-16y$$

Subtract 2y^{2} from both sides.

$$-z=-16-16y-2y^{2}$$

The equation is in standard form.

$$-z=-2y^{2}-16y-16$$

Divide both sides by -1.

$$\frac{-z}{-1}=\frac{-2y^{2}-16y-16}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$z=\frac{-2y^{2}-16y-16}{-1}$$

Divide -16-16y-2y^{2} by -1.

$$z=2y^{2}+16y+16$$

Solve for y (complex solution)

$$y=\frac{\sqrt{2z+32}}{2}-4$$

$$y=-\frac{\sqrt{2z+32}}{2}-4$$

Solve for y

$$y=\frac{\sqrt{2z+32}}{2}-4$$

$$y=-\frac{\sqrt{2z+32}}{2}-4\text{, }z\geq -16$$