-4.905t^2-24t+13=0

Solution for -4.905t^2-24t+13=0 equation:

-4.905t^2-24t+13=0

a = -4.905; b = -24; c = +13;
Δ = b2-4ac
Δ = -242-4·(-4.905)·13
Δ = 831.06
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{831.06}}{2*-4.905}=\frac{24-\sqrt{831.06}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+\sqrt{831.06}}{2*-4.905}=\frac{24+\sqrt{831.06}}{-9.81}
$