0=19.3t+-4.905t^2

Solution for 0=19.3t+-4.905t^2 equation:

0=19.3t+-4.905t^2

We move all terms to the left:

0-(19.3t+-4.905t^2)=0

We add all the numbers together, and all the variables

-(19.3t+-4.905t^2)=0

We use the square of the difference formula

-(19.3t-4.905t^2)=0

We get rid of parentheses

4.905t^2-19.3t=0

a = 4.905; b = -19.3; c = 0;
Δ = b2-4ac
Δ = -19.32-4·4.905·0
Δ = 372.49
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{-(-19.3)-\sqrt{372.49}}{2*4.905}=\frac{19.3-\sqrt{372.49}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19.3)+\sqrt{372.49}}{2*4.905}=\frac{19.3+\sqrt{372.49}}{9.81}
$