-16x^2-32x+240=0

Solution for -16x^2-32x+240=0 equation:

-16x^2-32x+240=0

a = -16; b = -32; c = +240;
Δ = b2-4ac
Δ = -322-4·(-16)·240
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-128}{2*-16}=\frac{-96}{-32}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+128}{2*-16}=\frac{160}{-32}
=-5
$