25y^2+35y-36=0

Solution for 25y^2+35y-36=0 equation:

25y^2+35y-36=0

a = 25; b = 35; c = -36;
Δ = b2-4ac
Δ = 352-4·25·(-36)
Δ = 4825
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:

$\sqrt{\Delta}=\sqrt{4825}=\sqrt{25*193}=\sqrt{25}*\sqrt{193}=5\sqrt{193}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-5\sqrt{193}}{2*25}=\frac{-35-5\sqrt{193}}{50}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+5\sqrt{193}}{2*25}=\frac{-35+5\sqrt{193}}{50}
$