2w^2+16w-1536=0

Solution for 2w^2+16w-1536=0 equation:

2w^2+16w-1536=0

a = 2; b = 16; c = -1536;
Δ = b2-4ac
Δ = 162-4·2·(-1536)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-112}{2*2}=\frac{-128}{4}
=-32
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+112}{2*2}=\frac{96}{4}
=24
$