0=-4.9t^2+13t+1

Solution for 0=-4.9t^2+13t+1 equation:

0=-4.9t^2+13t+1

We move all terms to the left:

0-(-4.9t^2+13t+1)=0

We add all the numbers together, and all the variables

-(-4.9t^2+13t+1)=0

We get rid of parentheses

4.9t^2-13t-1=0

a = 4.9; b = -13; c = -1;
Δ = b2-4ac
Δ = -132-4·4.9·(-1)
Δ = 188.6
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{-(-13)-\sqrt{188.6}}{2*4.9}=\frac{13-\sqrt{188.6}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{188.6}}{2*4.9}=\frac{13+\sqrt{188.6}}{9.8}
$