Solve n^2+2y^2+2x+2y+1=0

$${ n }^{ 2 } +2 { y }^{ 2 } +2x+2y+1 = 0$$

Solve for x

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

Steps for Solving Linear Equation

Subtract n^{2} from both sides. Anything subtracted from zero gives its negation.

$$2y^{2}+2x+2y+1=-n^{2}$$

Subtract 2y^{2} from both sides.

$$2x+2y+1=-n^{2}-2y^{2}$$

Subtract 2y from both sides.

$$2x+1=-n^{2}-2y^{2}-2y$$

Subtract 1 from both sides.

$$2x=-n^{2}-2y^{2}-2y-1$$

The equation is in standard form.

$$2x=-2y^{2}-2y-n^{2}-1$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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

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

Solve for n (complex solution)

$$n=-\sqrt{-2x-2y^{2}-2y-1}$$

$$n=\sqrt{-2x-2y^{2}-2y-1}$$

Solve for n

$$n=\sqrt{-2x-2y^{2}-2y-1}$$

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