4.9t^2+24t-2.5=0

Solution for 4.9t^2+24t-2.5=0 equation:

4.9t^2+24t-2.5=0

a = 4.9; b = 24; c = -2.5;
Δ = b2-4ac
Δ = 242-4·4.9·(-2.5)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-25}{2*4.9}=\frac{-49}{9.8}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+25}{2*4.9}=\frac{1}{9.8}
=1/9.8
$