0=0.13+2.6t-5t^2

Solution for 0=0.13+2.6t-5t^2 equation:

0=0.13+2.6t-5t^2

We move all terms to the left:

0-(0.13+2.6t-5t^2)=0

We add all the numbers together, and all the variables

-(0.13+2.6t-5t^2)=0

We get rid of parentheses

5t^2-2.6t-0.13=0

a = 5; b = -2.6; c = -0.13;
Δ = b2-4ac
Δ = -2.62-4·5·(-0.13)
Δ = 9.36
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{-(-2.6)-\sqrt{9.36}}{2*5}=\frac{2.6-\sqrt{9.36}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.6)+\sqrt{9.36}}{2*5}=\frac{2.6+\sqrt{9.36}}{10}
$