25y^2-40y+4=0

Solution for 25y^2-40y+4=0 equation:

25y^2-40y+4=0

a = 25; b = -40; c = +4;
Δ = b2-4ac
Δ = -402-4·25·4
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20\sqrt{3}}{2*25}=\frac{40-20\sqrt{3}}{50}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20\sqrt{3}}{2*25}=\frac{40+20\sqrt{3}}{50}
$