-16t^2+49.0237184t+6.3=0

Solution for -16t^2+49.0237184t+6.3=0 equation:

-16t^2+49.0237184t+6.3=0

a = -16; b = 49.0237184; c = +6.3;
Δ = b2-4ac
Δ = 49.02371842-4·(-16)·6.3
Δ = 2806.5249657625
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{-(49.0237184)-\sqrt{2806.5249657625}}{2*-16}=\frac{-49.0237184-\sqrt{2806.5249657625}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49.0237184)+\sqrt{2806.5249657625}}{2*-16}=\frac{-49.0237184+\sqrt{2806.5249657625}}{-32}
$