0=180-16t^2

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

0=180-16t^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(180-16t^2)=0

We get rid of parentheses

16t^2-180=0

a = 16; b = 0; c = -180;
Δ = b2-4ac
Δ = 02-4·16·(-180)
Δ = 11520
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:

$\sqrt{\Delta}=\sqrt{11520}=\sqrt{2304*5}=\sqrt{2304}*\sqrt{5}=48\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{5}}{2*16}=\frac{0-48\sqrt{5}}{32}
=-\frac{48\sqrt{5}}{32}
=-\frac{3\sqrt{5}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{5}}{2*16}=\frac{0+48\sqrt{5}}{32}
=\frac{48\sqrt{5}}{32}
=\frac{3\sqrt{5}}{2}
$