25y^2-20y+1=0

Solution for 25y^2-20y+1=0 equation:

25y^2-20y+1=0

a = 25; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·25·1
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{3}}{2*25}=\frac{20-10\sqrt{3}}{50}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{3}}{2*25}=\frac{20+10\sqrt{3}}{50}
$