2w^2+32w-6144=0

Solution for 2w^2+32w-6144=0 equation:

2w^2+32w-6144=0

a = 2; b = 32; c = -6144;
Δ = b2-4ac
Δ = 322-4·2·(-6144)
Δ = 50176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{50176}=224$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-224}{2*2}=\frac{-256}{4}
=-64
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+224}{2*2}=\frac{192}{4}
=48
$