0=11t-4.9t^2+24

Solution for 0=11t-4.9t^2+24 equation:

0=11t-4.9t^2+24

We move all terms to the left:

0-(11t-4.9t^2+24)=0

We add all the numbers together, and all the variables

-(11t-4.9t^2+24)=0

We get rid of parentheses

4.9t^2-11t-24=0

a = 4.9; b = -11; c = -24;
Δ = b2-4ac
Δ = -112-4·4.9·(-24)
Δ = 591.4
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{-(-11)-\sqrt{591.4}}{2*4.9}=\frac{11-\sqrt{591.4}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{591.4}}{2*4.9}=\frac{11+\sqrt{591.4}}{9.8}
$