4.9t^2+4.5t=0

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

4.9t^2+4.5t=0

a = 4.9; b = 4.5; c = 0;
Δ = b2-4ac
Δ = 4.52-4·4.9·0
Δ = 20.25
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{20.25}}{2*4.9}=\frac{-4.5-\sqrt{20.25}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.5)+\sqrt{20.25}}{2*4.9}=\frac{-4.5+\sqrt{20.25}}{9.8}
$