(t)=19.6t-4.9t^2

Solution for (t)=19.6t-4.9t^2 equation:

(t)=19.6t-4.9t^2

We move all terms to the left:

(t)-(19.6t-4.9t^2)=0

We get rid of parentheses

4.9t^2-19.6t+t=0

We add all the numbers together, and all the variables

4.9t^2-18.6t=0

a = 4.9; b = -18.6; c = 0;
Δ = b2-4ac
Δ = -18.62-4·4.9·0
Δ = 345.96
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{-(-18.6)-\sqrt{345.96}}{2*4.9}=\frac{18.6-\sqrt{345.96}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18.6)+\sqrt{345.96}}{2*4.9}=\frac{18.6+\sqrt{345.96}}{9.8}
$