0=-16t^2+32t+180

Solution for 0=-16t^2+32t+180 equation:

0=-16t^2+32t+180

We move all terms to the left:

0-(-16t^2+32t+180)=0

We add all the numbers together, and all the variables

-(-16t^2+32t+180)=0

We get rid of parentheses

16t^2-32t-180=0

a = 16; b = -32; c = -180;
Δ = b2-4ac
Δ = -322-4·16·(-180)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-112}{2*16}=\frac{-80}{32}
=-2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+112}{2*16}=\frac{144}{32}
=4+1/2
$