4.9t2-98t=0

Solution for 4.9t2-98t=0 equation:

4.9t^2-98t=0

a = 4.9; b = -98; c = 0;
Δ = b2-4ac
Δ = -982-4·4.9·0
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-98}{2*4.9}=\frac{0}{9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+98}{2*4.9}=\frac{196}{9.8}
=20
$