-28.42=0.3+5.5t+-4.9t2

Solution for -28.42=0.3+5.5t+-4.9t2 equation:

-28.42=0.3+5.5t+-4.9t^2

We move all terms to the left:

-28.42-(0.3+5.5t+-4.9t^2)=0

We use the square of the difference formula

-(0.3+5.5t-4.9t^2)-28.42=0

We get rid of parentheses

4.9t^2-5.5t-0.3-28.42=0

We add all the numbers together, and all the variables

4.9t^2-5.5t-28.72=0

a = 4.9; b = -5.5; c = -28.72;
Δ = b2-4ac
Δ = -5.52-4·4.9·(-28.72)
Δ = 593.162
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{-(-5.5)-\sqrt{593.162}}{2*4.9}=\frac{5.5-\sqrt{593.162}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5.5)+\sqrt{593.162}}{2*4.9}=\frac{5.5+\sqrt{593.162}}{9.8}
$