-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 = 6272
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:

$\sqrt{\Delta}=\sqrt{6272}=\sqrt{3136*2}=\sqrt{3136}*\sqrt{2}=56\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-56\sqrt{2}}{2*-16}=\frac{-80-56\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+56\sqrt{2}}{2*-16}=\frac{-80+56\sqrt{2}}{-32}
$