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

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

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