0=-16t^2+120t+100

Solution for 0=-16t^2+120t+100 equation:

0=-16t^2+120t+100

We move all terms to the left:

0-(-16t^2+120t+100)=0

We add all the numbers together, and all the variables

-(-16t^2+120t+100)=0

We get rid of parentheses

16t^2-120t-100=0

a = 16; b = -120; c = -100;
Δ = b2-4ac
Δ = -1202-4·16·(-100)
Δ = 20800
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:

$\sqrt{\Delta}=\sqrt{20800}=\sqrt{1600*13}=\sqrt{1600}*\sqrt{13}=40\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-40\sqrt{13}}{2*16}=\frac{120-40\sqrt{13}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+40\sqrt{13}}{2*16}=\frac{120+40\sqrt{13}}{32}
$