4.905t^2-12t-125=0

Solution for 4.905t^2-12t-125=0 equation:

4.905t^2-12t-125=0

a = 4.905; b = -12; c = -125;
Δ = b2-4ac
Δ = -122-4·4.905·(-125)
Δ = 2596.5
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{-(-12)-\sqrt{2596.5}}{2*4.905}=\frac{12-\sqrt{2596.5}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+\sqrt{2596.5}}{2*4.905}=\frac{12+\sqrt{2596.5}}{9.81}
$