Solve 2f(x)-5=(x-3)^2+2(3-x)

$$2 f ( x ) - 5 = ( x - 3 ) ^ { 2 } + 2 ( 3 - x )$$

Solve for f

$$f=\frac{x}{2}-4+\frac{10}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$2fx-5=x^{2}-6x+9+2\left(3-x\right)$$

Use the distributive property to multiply 2 by 3-x.

$$2fx-5=x^{2}-6x+9+6-2x$$

Add 9 and 6 to get 15.

$$2fx-5=x^{2}-6x+15-2x$$

Combine -6x and -2x to get -8x.

$$2fx-5=x^{2}-8x+15$$

Add 5 to both sides.

$$2fx=x^{2}-8x+15+5$$

Add 15 and 5 to get 20.

$$2fx=x^{2}-8x+20$$

The equation is in standard form.

$$2xf=x^{2}-8x+20$$

Divide both sides by 2x.

$$\frac{2xf}{2x}=\frac{x^{2}-8x+20}{2x}$$

Dividing by 2x undoes the multiplication by 2x.

$$f=\frac{x^{2}-8x+20}{2x}$$

Divide x^{2}-8x+20 by 2x.

$$f=\frac{x}{2}-4+\frac{10}{x}$$

Solve for x (complex solution)

$$x=-\sqrt{f^{2}+8f-4}+f+4$$

$$x=\sqrt{f^{2}+8f-4}+f+4$$

Solve for x

$$x=-\sqrt{f^{2}+8f-4}+f+4$$

$$x=\sqrt{f^{2}+8f-4}+f+4\text{, }f\geq 2\sqrt{5}-4\text{ or }f\leq -2\sqrt{5}-4$$