2w^2+3w-560=0

Solution for 2w^2+3w-560=0 equation:

2w^2+3w-560=0

a = 2; b = 3; c = -560;
Δ = b2-4ac
Δ = 32-4·2·(-560)
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-67}{2*2}=\frac{-70}{4}
=-17+1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+67}{2*2}=\frac{64}{4}
=16
$