Solve x^2+(9+y)x+16=0

$${ x }^{ 2 } +(9+y)x+16=0$$

Solve for y

$$y=-x-9-\frac{16}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 9+y by x.

$$x^{2}+9x+yx+16=0$$

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

$$9x+yx+16=-x^{2}$$

Subtract 9x from both sides.

$$yx+16=-x^{2}-9x$$

Subtract 16 from both sides.

$$yx=-x^{2}-9x-16$$

The equation is in standard form.

$$xy=-x^{2}-9x-16$$

Divide both sides by x.

$$\frac{xy}{x}=\frac{-x^{2}-9x-16}{x}$$

Dividing by x undoes the multiplication by x.

$$y=\frac{-x^{2}-9x-16}{x}$$

Divide -x^{2}-9x-16 by x.

$$y=-x-9-\frac{16}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(y+1\right)\left(y+17\right)}-y-9}{2}$$

$$x=\frac{-\sqrt{\left(y+1\right)\left(y+17\right)}-y-9}{2}$$

Solve for x

$$x=\frac{\sqrt{\left(y+1\right)\left(y+17\right)}-y-9}{2}$$

$$x=\frac{-\sqrt{\left(y+1\right)\left(y+17\right)}-y-9}{2}\text{, }y\leq -17\text{ or }y\geq -1$$