25w^2+20w-32=0

Solution for 25w^2+20w-32=0 equation:

25w^2+20w-32=0

a = 25; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·25·(-32)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-60}{2*25}=\frac{-80}{50}
=-1+3/5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+60}{2*25}=\frac{40}{50}
=4/5
$