4.9t^2+1.55t-2.8=0

Solution for 4.9t^2+1.55t-2.8=0 equation:

4.9t^2+1.55t-2.8=0

a = 4.9; b = 1.55; c = -2.8;
Δ = b2-4ac
Δ = 1.552-4·4.9·(-2.8)
Δ = 57.2825
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{-(1.55)-\sqrt{57.2825}}{2*4.9}=\frac{-1.55-\sqrt{57.2825}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.55)+\sqrt{57.2825}}{2*4.9}=\frac{-1.55+\sqrt{57.2825}}{9.8}
$