-16t^2+40t+380=0

Solution for -16t^2+40t+380=0 equation:

-16t^2+40t+380=0

a = -16; b = 40; c = +380;
Δ = b2-4ac
Δ = 402-4·(-16)·380
Δ = 25920
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:

$\sqrt{\Delta}=\sqrt{25920}=\sqrt{5184*5}=\sqrt{5184}*\sqrt{5}=72\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-72\sqrt{5}}{2*-16}=\frac{-40-72\sqrt{5}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+72\sqrt{5}}{2*-16}=\frac{-40+72\sqrt{5}}{-32}
$