0=11+2.5*t-4.9*t^2

Solution for 0=11+2.5*t-4.9*t^2 equation:

0=11+2.5t-4.9t^2

We move all terms to the left:

0-(11+2.5t-4.9t^2)=0

We add all the numbers together, and all the variables

-(11+2.5t-4.9t^2)=0

We get rid of parentheses

4.9t^2-2.5t-11=0

a = 4.9; b = -2.5; c = -11;
Δ = b2-4ac
Δ = -2.52-4·4.9·(-11)
Δ = 221.85
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{-(-2.5)-\sqrt{221.85}}{2*4.9}=\frac{2.5-\sqrt{221.85}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.5)+\sqrt{221.85}}{2*4.9}=\frac{2.5+\sqrt{221.85}}{9.8}
$