-16t^2+80t+40=0

Solution for -16t^2+80t+40=0 equation:

-16t^2+80t+40=0

a = -16; b = 80; c = +40;
Δ = b2-4ac
Δ = 802-4·(-16)·40
Δ = 8960
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:

$\sqrt{\Delta}=\sqrt{8960}=\sqrt{256*35}=\sqrt{256}*\sqrt{35}=16\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-16\sqrt{35}}{2*-16}=\frac{-80-16\sqrt{35}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+16\sqrt{35}}{2*-16}=\frac{-80+16\sqrt{35}}{-32}
$