15y^2+20y-25=0

Solution for 15y^2+20y-25=0 equation:

15y^2+20y-25=0

a = 15; b = 20; c = -25;
Δ = b2-4ac
Δ = 202-4·15·(-25)
Δ = 1900
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:

$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{19}}{2*15}=\frac{-20-10\sqrt{19}}{30}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{19}}{2*15}=\frac{-20+10\sqrt{19}}{30}
$