4.9t^2+4.5t-1.2=0

Solution for 4.9t^2+4.5t-1.2=0 equation:

4.9t^2+4.5t-1.2=0

a = 4.9; b = 4.5; c = -1.2;
Δ = b2-4ac
Δ = 4.52-4·4.9·(-1.2)
Δ = 43.77
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{-(4.5)-\sqrt{43.77}}{2*4.9}=\frac{-4.5-\sqrt{43.77}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.5)+\sqrt{43.77}}{2*4.9}=\frac{-4.5+\sqrt{43.77}}{9.8}
$