-16t^2+40t+8=0

Solution for -16t^2+40t+8=0 equation:

-16t^2+40t+8=0

a = -16; b = 40; c = +8;
Δ = b2-4ac
Δ = 402-4·(-16)·8
Δ = 2112
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:

$\sqrt{\Delta}=\sqrt{2112}=\sqrt{64*33}=\sqrt{64}*\sqrt{33}=8\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{33}}{2*-16}=\frac{-40-8\sqrt{33}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{33}}{2*-16}=\frac{-40+8\sqrt{33}}{-32}
$