-16t^2+20t+1100=0

Solution for -16t^2+20t+1100=0 equation:

-16t^2+20t+1100=0

a = -16; b = 20; c = +1100;
Δ = b2-4ac
Δ = 202-4·(-16)·1100
Δ = 70800
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:

$\sqrt{\Delta}=\sqrt{70800}=\sqrt{400*177}=\sqrt{400}*\sqrt{177}=20\sqrt{177}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{177}}{2*-16}=\frac{-20-20\sqrt{177}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{177}}{2*-16}=\frac{-20+20\sqrt{177}}{-32}
$