-16t^2+320t+20=0

Solution for -16t^2+320t+20=0 equation:

-16t^2+320t+20=0

a = -16; b = 320; c = +20;
Δ = b2-4ac
Δ = 3202-4·(-16)·20
Δ = 103680
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:

$\sqrt{\Delta}=\sqrt{103680}=\sqrt{20736*5}=\sqrt{20736}*\sqrt{5}=144\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-144\sqrt{5}}{2*-16}=\frac{-320-144\sqrt{5}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+144\sqrt{5}}{2*-16}=\frac{-320+144\sqrt{5}}{-32}
$