-4.9t^2+8t+2.2=0

Solution for -4.9t^2+8t+2.2=0 equation:

-4.9t^2+8t+2.2=0

a = -4.9; b = 8; c = +2.2;
Δ = b2-4ac
Δ = 82-4·(-4.9)·2.2
Δ = 107.12
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{107.12}}{2*-4.9}=\frac{-8-\sqrt{107.12}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+\sqrt{107.12}}{2*-4.9}=\frac{-8+\sqrt{107.12}}{-9.8}
$