-16t^2+128t+40=0

Solution for -16t^2+128t+40=0 equation:

-16t^2+128t+40=0

a = -16; b = 128; c = +40;
Δ = b2-4ac
Δ = 1282-4·(-16)·40
Δ = 18944
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{18944}=\sqrt{256*74}=\sqrt{256}*\sqrt{74}=16\sqrt{74}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-16\sqrt{74}}{2*-16}=\frac{-128-16\sqrt{74}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+16\sqrt{74}}{2*-16}=\frac{-128+16\sqrt{74}}{-32}
$