4.905t^2+39t-125=0

Solution for 4.905t^2+39t-125=0 equation:

4.905t^2+39t-125=0

a = 4.905; b = 39; c = -125;
Δ = b2-4ac
Δ = 392-4·4.905·(-125)
Δ = 3973.5
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{-(39)-\sqrt{3973.5}}{2*4.905}=\frac{-39-\sqrt{3973.5}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{3973.5}}{2*4.905}=\frac{-39+\sqrt{3973.5}}{9.81}
$