Solve -4x+y^2-32x+6y-91=0

$$-4x+ { y }^{ 2 } -32x+6y-91=0$$

Solve for x

$$x=\frac{\left(y-7\right)\left(y+13\right)}{36}$$

Steps for Solving Linear Equation

Combine -4x and -32x to get -36x.

$$-36x+y^{2}+6y-91=0$$

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

$$-36x+6y-91=-y^{2}$$

Subtract 6y from both sides.

$$-36x-91=-y^{2}-6y$$

Add 91 to both sides.

$$-36x=-y^{2}-6y+91$$

The equation is in standard form.

$$-36x=91-6y-y^{2}$$

Divide both sides by -36.

$$\frac{-36x}{-36}=\frac{\left(7-y\right)\left(y+13\right)}{-36}$$

Dividing by -36 undoes the multiplication by -36.

$$x=\frac{\left(7-y\right)\left(y+13\right)}{-36}$$

Divide \left(7-y\right)\left(13+y\right) by -36.

$$x=-\frac{\left(7-y\right)\left(y+13\right)}{36}$$

Solve for y (complex solution)

$$y=2\sqrt{9x+25}-3$$

$$y=-2\sqrt{9x+25}-3$$

Solve for y

$$y=2\sqrt{9x+25}-3$$

$$y=-2\sqrt{9x+25}-3\text{, }x\geq -\frac{25}{9}$$