Solve y^2+8y-4x=m

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

Solve for x

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

Steps for Solving Linear Equation

Subtract y^{2} from both sides.

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

Subtract 8y from both sides.

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

The equation is in standard form.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide m-y^{2}-8y by -4.

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

Solve for m

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

Solution Steps

Swap sides so that all variable terms are on the left hand side.

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