0=1.3t^2-10.28t-10.28

Solution for 0=1.3t^2-10.28t-10.28 equation:

0=1.3t^2-10.28t-10.28

We move all terms to the left:

0-(1.3t^2-10.28t-10.28)=0

We add all the numbers together, and all the variables

-(1.3t^2-10.28t-10.28)=0

We get rid of parentheses

-1.3t^2+10.28t+10.28=0

a = -1.3; b = 10.28; c = +10.28;
Δ = b2-4ac
Δ = 10.282-4·(-1.3)·10.28
Δ = 159.1344
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{-(10.28)-\sqrt{159.1344}}{2*-1.3}=\frac{-10.28-\sqrt{159.1344}}{-2.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10.28)+\sqrt{159.1344}}{2*-1.3}=\frac{-10.28+\sqrt{159.1344}}{-2.6}
$