Solve y^2-8x+6y+1=0

$$y ^ { 2 } - 8 x + 6 y + 1 = 0$$

Solve for x

$$x=\frac{y^{2}+6y+1}{8}$$

Steps for Solving Linear Equation

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

$$-8x+6y+1=-y^{2}$$

Subtract 6y from both sides.

$$-8x+1=-y^{2}-6y$$

Subtract 1 from both sides.

$$-8x=-y^{2}-6y-1$$

Divide both sides by -8.

$$\frac{-8x}{-8}=\frac{-y^{2}-6y-1}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x=\frac{-y^{2}-6y-1}{-8}$$

Divide -y^{2}-6y-1 by -8.

$$x=\frac{y^{2}}{8}+\frac{3y}{4}+\frac{1}{8}$$

Solve for y (complex solution)

$$y=2\sqrt{2\left(x+1\right)}-3$$

$$y=-2\sqrt{2\left(x+1\right)}-3$$

Solve for y

$$y=2\sqrt{2\left(x+1\right)}-3$$

$$y=-2\sqrt{2\left(x+1\right)}-3\text{, }x\geq -1$$