2w^2+13w-96=0

Solution for 2w^2+13w-96=0 equation:

2w^2+13w-96=0

a = 2; b = 13; c = -96;
Δ = b2-4ac
Δ = 132-4·2·(-96)
Δ = 937
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{937}}{2*2}=\frac{-13-\sqrt{937}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{937}}{2*2}=\frac{-13+\sqrt{937}}{4}
$