Solve x^2+4x-8y-20=0

$${ x }^{ 2 } +4x-8y-20=0$$

Solve for y

$$y=\frac{x^{2}+4x-20}{8}$$

Steps for Solving Linear Equation

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

$$4x-8y-20=-x^{2}$$

Subtract 4x from both sides.

$$-8y-20=-x^{2}-4x$$

Add 20 to both sides.

$$-8y=-x^{2}-4x+20$$

The equation is in standard form.

$$-8y=20-4x-x^{2}$$

Divide both sides by -8.

$$\frac{-8y}{-8}=\frac{20-4x-x^{2}}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$y=\frac{20-4x-x^{2}}{-8}$$

Divide -x^{2}-4x+20 by -8.

$$y=\frac{x^{2}}{8}+\frac{x}{2}-\frac{5}{2}$$

Solve for x (complex solution)

$$x=2\sqrt{2y+6}-2$$

$$x=-2\sqrt{2y+6}-2$$

Solve for x

$$x=2\sqrt{2y+6}-2$$

$$x=-2\sqrt{2y+6}-2\text{, }y\geq -3$$