Solve x^2+8x-4y+36=0

$${ x }^{ 2 } +8x-4y+36=0$$

Solve for y

$$y=\frac{x^{2}+8x+36}{4}$$

Steps for Solving Linear Equation

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

$$8x-4y+36=-x^{2}$$

Subtract 8x from both sides.

$$-4y+36=-x^{2}-8x$$

Subtract 36 from both sides.

$$-4y=-x^{2}-8x-36$$

Divide both sides by -4.

$$\frac{-4y}{-4}=\frac{-x^{2}-8x-36}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$y=\frac{-x^{2}-8x-36}{-4}$$

Divide -x^{2}-8x-36 by -4.

$$y=\frac{x^{2}}{4}+2x+9$$

Solve for x (complex solution)

$$x=2\sqrt{y-5}-4$$

$$x=-2\sqrt{y-5}-4$$

Solve for x

$$x=2\sqrt{y-5}-4$$

$$x=-2\sqrt{y-5}-4\text{, }y\geq 5$$