0=6.6(t)+4.9(t^2)

Solution for 0=6.6(t)+4.9(t^2) equation:

0=6.6(t)+4.9(t^2)

We move all terms to the left:

0-(6.6(t)+4.9(t^2))=0

We add all the numbers together, and all the variables

-(6.6t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-6.6t=0

a = -4.9; b = -6.6; c = 0;
Δ = b2-4ac
Δ = -6.62-4·(-4.9)·0
Δ = 43.56
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{-(-6.6)-\sqrt{43.56}}{2*-4.9}=\frac{6.6-\sqrt{43.56}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6.6)+\sqrt{43.56}}{2*-4.9}=\frac{6.6+\sqrt{43.56}}{-9.8}
$