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

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

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

a = -16; b = 128; c = +2;
Δ = b2-4ac
Δ = 1282-4·(-16)·2
Δ = 16512
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:

$\sqrt{\Delta}=\sqrt{16512}=\sqrt{64*258}=\sqrt{64}*\sqrt{258}=8\sqrt{258}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-8\sqrt{258}}{2*-16}=\frac{-128-8\sqrt{258}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+8\sqrt{258}}{2*-16}=\frac{-128+8\sqrt{258}}{-32}
$