4.9t^2-24.5t+4.9=0

Solution for 4.9t^2-24.5t+4.9=0 equation:

4.9t^2-24.5t+4.9=0

a = 4.9; b = -24.5; c = +4.9;
Δ = b2-4ac
Δ = -24.52-4·4.9·4.9
Δ = 504.21
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{-(-24.5)-\sqrt{504.21}}{2*4.9}=\frac{24.5-\sqrt{504.21}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24.5)+\sqrt{504.21}}{2*4.9}=\frac{24.5+\sqrt{504.21}}{9.8}
$