0=4.9t^2+120t

Solution for 0=4.9t^2+120t equation:

0=4.9t^2+120t

We move all terms to the left:

0-(4.9t^2+120t)=0

We add all the numbers together, and all the variables

-(4.9t^2+120t)=0

We get rid of parentheses

-4.9t^2-120t=0

a = -4.9; b = -120; c = 0;
Δ = b2-4ac
Δ = -1202-4·(-4.9)·0
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-120}{2*-4.9}=\frac{0}{-9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+120}{2*-4.9}=\frac{240}{-9.8}
=-24+2.6666666666667/5.4444444444445
$