16w^2+64w-128=0

Solution for 16w^2+64w-128=0 equation:

16w^2+64w-128=0

a = 16; b = 64; c = -128;
Δ = b2-4ac
Δ = 642-4·16·(-128)
Δ = 12288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12288}=\sqrt{4096*3}=\sqrt{4096}*\sqrt{3}=64\sqrt{3}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-64\sqrt{3}}{2*16}=\frac{-64-64\sqrt{3}}{32}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64\sqrt{3}}{2*16}=\frac{-64+64\sqrt{3}}{32}
$