-16t^2+80t+10=0

Solution for -16t^2+80t+10=0 equation:

-16t^2+80t+10=0

a = -16; b = 80; c = +10;
Δ = b2-4ac
Δ = 802-4·(-16)·10
Δ = 7040
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:

$\sqrt{\Delta}=\sqrt{7040}=\sqrt{64*110}=\sqrt{64}*\sqrt{110}=8\sqrt{110}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-8\sqrt{110}}{2*-16}=\frac{-80-8\sqrt{110}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+8\sqrt{110}}{2*-16}=\frac{-80+8\sqrt{110}}{-32}
$