28=26+9t-4.9t^2

Solution for 28=26+9t-4.9t^2 equation:

28=26+9t-4.9t^2

We move all terms to the left:

28-(26+9t-4.9t^2)=0

We get rid of parentheses

4.9t^2-9t-26+28=0

We add all the numbers together, and all the variables

4.9t^2-9t+2=0

a = 4.9; b = -9; c = +2;
Δ = b2-4ac
Δ = -92-4·4.9·2
Δ = 41.8
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{-(-9)-\sqrt{41.8}}{2*4.9}=\frac{9-\sqrt{41.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{41.8}}{2*4.9}=\frac{9+\sqrt{41.8}}{9.8}
$