-16t^2+128t+12=0

Solution for -16t^2+128t+12=0 equation:

-16t^2+128t+12=0

a = -16; b = 128; c = +12;
Δ = b2-4ac
Δ = 1282-4·(-16)·12
Δ = 17152
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:

$\sqrt{\Delta}=\sqrt{17152}=\sqrt{256*67}=\sqrt{256}*\sqrt{67}=16\sqrt{67}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-16\sqrt{67}}{2*-16}=\frac{-128-16\sqrt{67}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+16\sqrt{67}}{2*-16}=\frac{-128+16\sqrt{67}}{-32}
$