-16t^2+80t+2=0

Solution for -16t^2+80t+2=0 equation:

-16t^2+80t+2=0

a = -16; b = 80; c = +2;
Δ = b2-4ac
Δ = 802-4·(-16)·2
Δ = 6528
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:

$\sqrt{\Delta}=\sqrt{6528}=\sqrt{64*102}=\sqrt{64}*\sqrt{102}=8\sqrt{102}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-8\sqrt{102}}{2*-16}=\frac{-80-8\sqrt{102}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+8\sqrt{102}}{2*-16}=\frac{-80+8\sqrt{102}}{-32}
$