-16t^2-96t+160=0

Solution for -16t^2-96t+160=0 equation:

-16t^2-96t+160=0

a = -16; b = -96; c = +160;
Δ = b2-4ac
Δ = -962-4·(-16)·160
Δ = 19456
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:

$\sqrt{\Delta}=\sqrt{19456}=\sqrt{1024*19}=\sqrt{1024}*\sqrt{19}=32\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-32\sqrt{19}}{2*-16}=\frac{96-32\sqrt{19}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+32\sqrt{19}}{2*-16}=\frac{96+32\sqrt{19}}{-32}
$