-4.905t^2-2.8t=0

Solution for -4.905t^2-2.8t=0 equation:

-4.905t^2-2.8t=0

a = -4.905; b = -2.8; c = 0;
Δ = b2-4ac
Δ = -2.82-4·(-4.905)·0
Δ = 7.84
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{-(-2.8)-\sqrt{7.84}}{2*-4.905}=\frac{2.8-\sqrt{7.84}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.8)+\sqrt{7.84}}{2*-4.905}=\frac{2.8+\sqrt{7.84}}{-9.81}
$