-16t^2+9t+180=0

Solution for -16t^2+9t+180=0 equation:

-16t^2+9t+180=0

a = -16; b = 9; c = +180;
Δ = b2-4ac
Δ = 92-4·(-16)·180
Δ = 11601
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:

$\sqrt{\Delta}=\sqrt{11601}=\sqrt{9*1289}=\sqrt{9}*\sqrt{1289}=3\sqrt{1289}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{1289}}{2*-16}=\frac{-9-3\sqrt{1289}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{1289}}{2*-16}=\frac{-9+3\sqrt{1289}}{-32}
$