0=-10t^2+40t+120

Solution for 0=-10t^2+40t+120 equation:

0=-10t^2+40t+120

We move all terms to the left:

0-(-10t^2+40t+120)=0

We add all the numbers together, and all the variables

-(-10t^2+40t+120)=0

We get rid of parentheses

10t^2-40t-120=0

a = 10; b = -40; c = -120;
Δ = b2-4ac
Δ = -402-4·10·(-120)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-80}{2*10}=\frac{-40}{20}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+80}{2*10}=\frac{120}{20}
=6
$