Solve 2x^2-3x+1=f(x)

$$2 x ^ { 2 } - 3 x + 1 = f ( x )$$

Solve for f

$$f=2x-3+\frac{1}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$fx=2x^{2}-3x+1$$

The equation is in standard form.

$$xf=2x^{2}-3x+1$$

Divide both sides by x.

$$\frac{xf}{x}=\frac{\left(x-1\right)\left(2x-1\right)}{x}$$

Dividing by x undoes the multiplication by x.

$$f=\frac{\left(x-1\right)\left(2x-1\right)}{x}$$

Divide \left(-1+x\right)\left(-1+2x\right) by x.

$$f=2x-3+\frac{1}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{f^{2}+6f+1}+f+3}{4}$$

$$x=\frac{-\sqrt{f^{2}+6f+1}+f+3}{4}$$

Solve for x

$$x=\frac{\sqrt{f^{2}+6f+1}+f+3}{4}$$

$$x=\frac{-\sqrt{f^{2}+6f+1}+f+3}{4}\text{, }f\geq 2\sqrt{2}-3\text{ or }f\leq -2\sqrt{2}-3$$