0=4.9t^2+10t-120

Solution for 0=4.9t^2+10t-120 equation:

0=4.9t^2+10t-120

We move all terms to the left:

0-(4.9t^2+10t-120)=0

We add all the numbers together, and all the variables

-(4.9t^2+10t-120)=0

We get rid of parentheses

-4.9t^2-10t+120=0

a = -4.9; b = -10; c = +120;
Δ = b2-4ac
Δ = -102-4·(-4.9)·120
Δ = 2452
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:

$\sqrt{\Delta}=\sqrt{2452}=\sqrt{4*613}=\sqrt{4}*\sqrt{613}=2\sqrt{613}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{613}}{2*-4.9}=\frac{10-2\sqrt{613}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{613}}{2*-4.9}=\frac{10+2\sqrt{613}}{-9.8}
$