-4.905t^2-120t+125=0

Solution for -4.905t^2-120t+125=0 equation:

-4.905t^2-120t+125=0

a = -4.905; b = -120; c = +125;
Δ = b2-4ac
Δ = -1202-4·(-4.905)·125
Δ = 16852.5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-\sqrt{16852.5}}{2*-4.905}=\frac{120-\sqrt{16852.5}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+\sqrt{16852.5}}{2*-4.905}=\frac{120+\sqrt{16852.5}}{-9.81}
$