4,9t^2+13t-24=0

Solution for 4,9t^2+13t-24=0 equation:

4.9t^2+13t-24=0

a = 4.9; b = 13; c = -24;
Δ = b2-4ac
Δ = 132-4·4.9·(-24)
Δ = 639.4
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{-(13)-\sqrt{639.4}}{2*4.9}=\frac{-13-\sqrt{639.4}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{639.4}}{2*4.9}=\frac{-13+\sqrt{639.4}}{9.8}
$