0=-16t^2+30t+180

Solution for 0=-16t^2+30t+180 equation:

0=-16t^2+30t+180

We move all terms to the left:

0-(-16t^2+30t+180)=0

We add all the numbers together, and all the variables

-(-16t^2+30t+180)=0

We get rid of parentheses

16t^2-30t-180=0

a = 16; b = -30; c = -180;
Δ = b2-4ac
Δ = -302-4·16·(-180)
Δ = 12420
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:

$\sqrt{\Delta}=\sqrt{12420}=\sqrt{36*345}=\sqrt{36}*\sqrt{345}=6\sqrt{345}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-6\sqrt{345}}{2*16}=\frac{30-6\sqrt{345}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+6\sqrt{345}}{2*16}=\frac{30+6\sqrt{345}}{32}
$