4.9t^2-4t-1.8=0

Solution for 4.9t^2-4t-1.8=0 equation:

4.9t^2-4t-1.8=0

a = 4.9; b = -4; c = -1.8;
Δ = b2-4ac
Δ = -42-4·4.9·(-1.8)
Δ = 51.28
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{-(-4)-\sqrt{51.28}}{2*4.9}=\frac{4-\sqrt{51.28}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{51.28}}{2*4.9}=\frac{4+\sqrt{51.28}}{9.8}
$