-16t^2-40t+10=0

Solution for -16t^2-40t+10=0 equation:

-16t^2-40t+10=0

a = -16; b = -40; c = +10;
Δ = b2-4ac
Δ = -402-4·(-16)·10
Δ = 2240
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:

$\sqrt{\Delta}=\sqrt{2240}=\sqrt{64*35}=\sqrt{64}*\sqrt{35}=8\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{35}}{2*-16}=\frac{40-8\sqrt{35}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{35}}{2*-16}=\frac{40+8\sqrt{35}}{-32}
$