4.9t^2+1.55t-4.9=0

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

4.9t^2+1.55t-4.9=0

a = 4.9; b = 1.55; c = -4.9;
Δ = b2-4ac
Δ = 1.552-4·4.9·(-4.9)
Δ = 98.4425
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{98.4425}}{2*4.9}=\frac{-1.55-\sqrt{98.4425}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.55)+\sqrt{98.4425}}{2*4.9}=\frac{-1.55+\sqrt{98.4425}}{9.8}
$