20y^2-10y+1=0

Solution for 20y^2-10y+1=0 equation:

20y^2-10y+1=0

a = 20; b = -10; c = +1;
Δ = b2-4ac
Δ = -102-4·20·1
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{5}}{2*20}=\frac{10-2\sqrt{5}}{40}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{5}}{2*20}=\frac{10+2\sqrt{5}}{40}
$