0=16t^2+21120t

Solution for 0=16t^2+21120t equation:

0=16t^2+21120t

We move all terms to the left:

0-(16t^2+21120t)=0

We add all the numbers together, and all the variables

-(16t^2+21120t)=0

We get rid of parentheses

-16t^2-21120t=0

a = -16; b = -21120; c = 0;
Δ = b2-4ac
Δ = -211202-4·(-16)·0
Δ = 446054400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{446054400}=21120$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21120)-21120}{2*-16}=\frac{0}{-32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21120)+21120}{2*-16}=\frac{42240}{-32}
=-1320
$