4.905t^2+24t+7=0

Solution for 4.905t^2+24t+7=0 equation:

4.905t^2+24t+7=0

a = 4.905; b = 24; c = +7;
Δ = b2-4ac
Δ = 242-4·4.905·7
Δ = 438.66
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)-\sqrt{438.66}}{2*4.905}=\frac{-24-\sqrt{438.66}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+\sqrt{438.66}}{2*4.905}=\frac{-24+\sqrt{438.66}}{9.81}
$