Solve f(x)=-x^2+9quadg(x)=6-2x

$$f ( x ) = - x ^ { 2 } + 9 \quad g ( x ) = 6 - 2 x$$

Solve for g

$$g=\frac{x}{9}-\frac{2}{9}+\frac{2}{3x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Add x^{2} to both sides.

$$9gx=6-2x+x^{2}$$

The equation is in standard form.

$$9xg=x^{2}-2x+6$$

Divide both sides by 9x.

$$\frac{9xg}{9x}=\frac{x^{2}-2x+6}{9x}$$

Dividing by 9x undoes the multiplication by 9x.

$$g=\frac{x^{2}-2x+6}{9x}$$

Divide 6-2x+x^{2} by 9x.

$$g=\frac{x}{9}-\frac{2}{9}+\frac{2}{3x}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{81g^{2}+36g-20}}{2}+\frac{9g}{2}+1$$

$$x=\frac{\sqrt{81g^{2}+36g-20}}{2}+\frac{9g}{2}+1$$

Solve for x

$$x=-\frac{\sqrt{81g^{2}+36g-20}}{2}+\frac{9g}{2}+1$$

$$x=\frac{\sqrt{81g^{2}+36g-20}}{2}+\frac{9g}{2}+1\text{, }g\geq \frac{2\sqrt{6}-2}{9}\text{ or }g\leq \frac{-2\sqrt{6}-2}{9}$$