0=-4.9t^2+2.5t+13

Solution for 0=-4.9t^2+2.5t+13 equation:

0=-4.9t^2+2.5t+13

We move all terms to the left:

0-(-4.9t^2+2.5t+13)=0

We add all the numbers together, and all the variables

-(-4.9t^2+2.5t+13)=0

We get rid of parentheses

4.9t^2-2.5t-13=0

a = 4.9; b = -2.5; c = -13;
Δ = b2-4ac
Δ = -2.52-4·4.9·(-13)
Δ = 261.05
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{-(-2.5)-\sqrt{261.05}}{2*4.9}=\frac{2.5-\sqrt{261.05}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.5)+\sqrt{261.05}}{2*4.9}=\frac{2.5+\sqrt{261.05}}{9.8}
$