0=21.48t+4.9t^2

Solution for 0=21.48t+4.9t^2 equation:

0=21.48t+4.9t^2

We move all terms to the left:

0-(21.48t+4.9t^2)=0

We add all the numbers together, and all the variables

-(21.48t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-21.48t=0

a = -4.9; b = -21.48; c = 0;
Δ = b2-4ac
Δ = -21.482-4·(-4.9)·0
Δ = 461.3904
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{-(-21.48)-\sqrt{461.3904}}{2*-4.9}=\frac{21.48-\sqrt{461.3904}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21.48)+\sqrt{461.3904}}{2*-4.9}=\frac{21.48+\sqrt{461.3904}}{-9.8}
$