Solve x^2+4x+8y-4=0

$${ x }^{ 2 } +4x+8y-4=0$$

Solve for y

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

Steps for Solving Linear Equation

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

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

Subtract 4x from both sides.

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

Add 4 to both sides.

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

The equation is in standard form.

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

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

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

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

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