Solve x^2+y2-8x+10y+5=0

$${ x }^{ 2 } +y2-8x+10y+5=0$$

Solve for y

$$y=-\frac{x^{2}}{10}+\frac{4x}{5}-\frac{y_{2}}{10}-\frac{1}{2}$$

Steps for Solving Linear Equation

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

$$y_{2}-8x+10y+5=-x^{2}$$

Subtract y_{2} from both sides.

$$-8x+10y+5=-x^{2}-y_{2}$$

Add 8x to both sides.

$$10y+5=-x^{2}-y_{2}+8x$$

Subtract 5 from both sides.

$$10y=-x^{2}-y_{2}+8x-5$$

The equation is in standard form.

$$10y=-x^{2}+8x-y_{2}-5$$

Divide both sides by 10.

$$\frac{10y}{10}=\frac{-x^{2}+8x-y_{2}-5}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$y=\frac{-x^{2}+8x-y_{2}-5}{10}$$

Divide -x^{2}-y_{2}+8x-5 by 10.

$$y=-\frac{x^{2}}{10}+\frac{4x}{5}-\frac{y_{2}}{10}-\frac{1}{2}$$

Solve for x (complex solution)

$$x=\sqrt{11-y_{2}-10y}+4$$

$$x=-\sqrt{11-y_{2}-10y}+4$$

Solve for x

$$x=\sqrt{11-y_{2}-10y}+4$$

$$x=-\sqrt{11-y_{2}-10y}+4\text{, }y_{2}\leq 11-10y$$