4.9t^2+(-8t)+2.8=0

Solution for 4.9t^2+(-8t)+2.8=0 equation:

4.9t^2+(-8t)+2.8=0

We get rid of parentheses

4.9t^2-8t+2.8=0

a = 4.9; b = -8; c = +2.8;
Δ = b2-4ac
Δ = -82-4·4.9·2.8
Δ = 9.12
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{-(-8)-\sqrt{9.12}}{2*4.9}=\frac{8-\sqrt{9.12}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+\sqrt{9.12}}{2*4.9}=\frac{8+\sqrt{9.12}}{9.8}
$