Solve (x-4)^2+3=f(x)

$$( x - 4 ) ^ { 2 } + 3 = f ( x )$$

Solve for f

$$f=x-8+\frac{19}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16+3=fx$$

Add 16 and 3 to get 19.

$$x^{2}-8x+19=fx$$

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

$$fx=x^{2}-8x+19$$

The equation is in standard form.

$$xf=x^{2}-8x+19$$

Divide both sides by x.

$$\frac{xf}{x}=\frac{x^{2}-8x+19}{x}$$

Dividing by x undoes the multiplication by x.

$$f=\frac{x^{2}-8x+19}{x}$$

Divide x^{2}-8x+19 by x.

$$f=x-8+\frac{19}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{f^{2}+16f-12}+f+8}{2}$$

$$x=\frac{-\sqrt{f^{2}+16f-12}+f+8}{2}$$

Solve for x

$$x=\frac{\sqrt{f^{2}+16f-12}+f+8}{2}$$

$$x=\frac{-\sqrt{f^{2}+16f-12}+f+8}{2}\text{, }f\geq 2\sqrt{19}-8\text{ or }f\leq -2\sqrt{19}-8$$