4.9t^2-9.8t=0

Solution for 4.9t^2-9.8t=0 equation:

4.9t^2-9.8t=0

a = 4.9; b = -9.8; c = 0;
Δ = b2-4ac
Δ = -9.82-4·4.9·0
Δ = 96.04
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{-(-9.8)-\sqrt{96.04}}{2*4.9}=\frac{9.8-\sqrt{96.04}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9.8)+\sqrt{96.04}}{2*4.9}=\frac{9.8+\sqrt{96.04}}{9.8}
$