-16t2+36t+160=0

Solution for -16t2+36t+160=0 equation:

-16t^2+36t+160=0

a = -16; b = 36; c = +160;
Δ = b2-4ac
Δ = 362-4·(-16)·160
Δ = 11536
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:

$\sqrt{\Delta}=\sqrt{11536}=\sqrt{16*721}=\sqrt{16}*\sqrt{721}=4\sqrt{721}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{721}}{2*-16}=\frac{-36-4\sqrt{721}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{721}}{2*-16}=\frac{-36+4\sqrt{721}}{-32}
$