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

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

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

a = -16; b = 16; c = +320;
Δ = b2-4ac
Δ = 162-4·(-16)·320
Δ = 20736
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{20736}=144$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-144}{2*-16}=\frac{-160}{-32}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+144}{2*-16}=\frac{128}{-32}
=-4
$