4.9t^2+39t-125=0

Solution for 4.9t^2+39t-125=0 equation:

4.9t^2+39t-125=0

a = 4.9; b = 39; c = -125;
Δ = b2-4ac
Δ = 392-4·4.9·(-125)
Δ = 3971
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3971}=\sqrt{361*11}=\sqrt{361}*\sqrt{11}=19\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-19\sqrt{11}}{2*4.9}=\frac{-39-19\sqrt{11}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+19\sqrt{11}}{2*4.9}=\frac{-39+19\sqrt{11}}{9.8}
$