4w^2+24w-64=0

Solution for 4w^2+24w-64=0 equation:

4w^2+24w-64=0

a = 4; b = 24; c = -64;
Δ = b2-4ac
Δ = 242-4·4·(-64)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-40}{2*4}=\frac{-64}{8}
=-8
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+40}{2*4}=\frac{16}{8}
=2
$