25y^2+20y-4=0

Solution for 25y^2+20y-4=0 equation:

25y^2+20y-4=0

a = 25; b = 20; c = -4;
Δ = b2-4ac
Δ = 202-4·25·(-4)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{2}}{2*25}=\frac{-20-20\sqrt{2}}{50}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{2}}{2*25}=\frac{-20+20\sqrt{2}}{50}
$