0=-16t^2+45t+180

Solution for 0=-16t^2+45t+180 equation:

0=-16t^2+45t+180

We move all terms to the left:

0-(-16t^2+45t+180)=0

We add all the numbers together, and all the variables

-(-16t^2+45t+180)=0

We get rid of parentheses

16t^2-45t-180=0

a = 16; b = -45; c = -180;
Δ = b2-4ac
Δ = -452-4·16·(-180)
Δ = 13545
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:

$\sqrt{\Delta}=\sqrt{13545}=\sqrt{9*1505}=\sqrt{9}*\sqrt{1505}=3\sqrt{1505}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-3\sqrt{1505}}{2*16}=\frac{45-3\sqrt{1505}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+3\sqrt{1505}}{2*16}=\frac{45+3\sqrt{1505}}{32}
$