2f^2-4f+1=2f

Solution for 2f^2-4f+1=2f equation:

2f^2-4f+1=2f

We move all terms to the left:

2f^2-4f+1-(2f)=0

We add all the numbers together, and all the variables

2f^2-6f+1=0

a = 2; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·2·1
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{7}}{2*2}=\frac{6-2\sqrt{7}}{4}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{7}}{2*2}=\frac{6+2\sqrt{7}}{4}
$