0=-16t^2+120t+20

Solution for 0=-16t^2+120t+20 equation:

0=-16t^2+120t+20

We move all terms to the left:

0-(-16t^2+120t+20)=0

We add all the numbers together, and all the variables

-(-16t^2+120t+20)=0

We get rid of parentheses

16t^2-120t-20=0

a = 16; b = -120; c = -20;
Δ = b2-4ac
Δ = -1202-4·16·(-20)
Δ = 15680
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:

$\sqrt{\Delta}=\sqrt{15680}=\sqrt{3136*5}=\sqrt{3136}*\sqrt{5}=56\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-56\sqrt{5}}{2*16}=\frac{120-56\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+56\sqrt{5}}{2*16}=\frac{120+56\sqrt{5}}{32}
$