-16t^2-24t+180=0

Solution for -16t^2-24t+180=0 equation:

-16t^2-24t+180=0

a = -16; b = -24; c = +180;
Δ = b2-4ac
Δ = -242-4·(-16)·180
Δ = 12096
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:

$\sqrt{\Delta}=\sqrt{12096}=\sqrt{576*21}=\sqrt{576}*\sqrt{21}=24\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24\sqrt{21}}{2*-16}=\frac{24-24\sqrt{21}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24\sqrt{21}}{2*-16}=\frac{24+24\sqrt{21}}{-32}
$