25y^2-252y+160=0

Solution for 25y^2-252y+160=0 equation:

25y^2-252y+160=0

a = 25; b = -252; c = +160;
Δ = b2-4ac
Δ = -2522-4·25·160
Δ = 47504
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:

$\sqrt{\Delta}=\sqrt{47504}=\sqrt{16*2969}=\sqrt{16}*\sqrt{2969}=4\sqrt{2969}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-252)-4\sqrt{2969}}{2*25}=\frac{252-4\sqrt{2969}}{50}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-252)+4\sqrt{2969}}{2*25}=\frac{252+4\sqrt{2969}}{50}
$