0=-16t^2-13t+180

Solution for 0=-16t^2-13t+180 equation:

0=-16t^2-13t+180

We move all terms to the left:

0-(-16t^2-13t+180)=0

We add all the numbers together, and all the variables

-(-16t^2-13t+180)=0

We get rid of parentheses

16t^2+13t-180=0

a = 16; b = 13; c = -180;
Δ = b2-4ac
Δ = 132-4·16·(-180)
Δ = 11689
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{11689}}{2*16}=\frac{-13-\sqrt{11689}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{11689}}{2*16}=\frac{-13+\sqrt{11689}}{32}
$