-16t^2-12t+180=0

Solution for -16t^2-12t+180=0 equation:

-16t^2-12t+180=0

a = -16; b = -12; c = +180;
Δ = b2-4ac
Δ = -122-4·(-16)·180
Δ = 11664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11664}=108$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-108}{2*-16}=\frac{-96}{-32}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+108}{2*-16}=\frac{120}{-32}
=-3+3/4
$