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

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

Solve for x

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

Steps for Solving Linear Equation

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

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

Add y^{2} to both sides. Anything plus zero gives itself.

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

Add y to both sides.

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

Subtract 4 from both sides.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide y^{2}+y-4 by -1.

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

Solve for y (complex solution)

$$y=\frac{-\sqrt{17-4x}-1}{2}$$

$$y=\frac{\sqrt{17-4x}-1}{2}$$

Solve for y

$$y=\frac{-\sqrt{17-4x}-1}{2}$$

$$y=\frac{\sqrt{17-4x}-1}{2}\text{, }x\leq \frac{17}{4}$$