0=11t-4.9t^2

Solution for 0=11t-4.9t^2 equation:

0=11t-4.9t^2

We move all terms to the left:

0-(11t-4.9t^2)=0

We add all the numbers together, and all the variables

-(11t-4.9t^2)=0

We get rid of parentheses

4.9t^2-11t=0

a = 4.9; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·4.9·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*4.9}=\frac{0}{9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*4.9}=\frac{22}{9.8}
=2+2.4/9.8
$