-2t2+24t+128=0

Solution for -2t2+24t+128=0 equation:

-2t^2+24t+128=0

a = -2; b = 24; c = +128;
Δ = b2-4ac
Δ = 242-4·(-2)·128
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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