-16t^2+80t+8=0

Solution for -16t^2+80t+8=0 equation:

-16t^2+80t+8=0

a = -16; b = 80; c = +8;
Δ = b2-4ac
Δ = 802-4·(-16)·8
Δ = 6912
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:

$\sqrt{\Delta}=\sqrt{6912}=\sqrt{2304*3}=\sqrt{2304}*\sqrt{3}=48\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-48\sqrt{3}}{2*-16}=\frac{-80-48\sqrt{3}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+48\sqrt{3}}{2*-16}=\frac{-80+48\sqrt{3}}{-32}
$