-16t^2+52t+220=0

Solution for -16t^2+52t+220=0 equation:

-16t^2+52t+220=0

a = -16; b = 52; c = +220;
Δ = b2-4ac
Δ = 522-4·(-16)·220
Δ = 16784
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:

$\sqrt{\Delta}=\sqrt{16784}=\sqrt{16*1049}=\sqrt{16}*\sqrt{1049}=4\sqrt{1049}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-4\sqrt{1049}}{2*-16}=\frac{-52-4\sqrt{1049}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+4\sqrt{1049}}{2*-16}=\frac{-52+4\sqrt{1049}}{-32}
$