Solve left(y+4right)^2=x^2+4+4x+y^2+9+6y

$${ \left(y+4 \right) }^{ 2 } = { x }^{ 2 } +4+4x+ { y }^{ 2 } +9+6y$$

Solve for y

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

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}.

$$y^{2}+8y+16=x^{2}+4+4x+y^{2}+9+6y$$

Add 4 and 9 to get 13.

$$y^{2}+8y+16=x^{2}+13+4x+y^{2}+6y$$

Subtract y^{2} from both sides.

$$y^{2}+8y+16-y^{2}=x^{2}+13+4x+6y$$

Combine y^{2} and -y^{2} to get 0.

$$8y+16=x^{2}+13+4x+6y$$

Subtract 6y from both sides.

$$8y+16-6y=x^{2}+13+4x$$

Combine 8y and -6y to get 2y.

$$2y+16=x^{2}+13+4x$$

Subtract 16 from both sides.

$$2y=x^{2}+13+4x-16$$

Subtract 16 from 13 to get -3.

$$2y=x^{2}-3+4x$$

The equation is in standard form.

$$2y=x^{2}+4x-3$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide x^{2}-3+4x by 2.

$$y=\frac{x^{2}}{2}+2x-\frac{3}{2}$$

Solve for x (complex solution)

$$x=\sqrt{2y+7}-2$$

$$x=-\sqrt{2y+7}-2$$

Solve for x

$$x=\sqrt{2y+7}-2$$

$$x=-\sqrt{2y+7}-2\text{, }y\geq -\frac{7}{2}$$