2q^2-4q+1=0

Solution for 2q^2-4q+1=0 equation:

2q^2-4q+1=0

a = 2; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·2·1
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{2}}{2*2}=\frac{4-2\sqrt{2}}{4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{2}}{2*2}=\frac{4+2\sqrt{2}}{4}
$