0=-0.8t^2+10t+2

Solution for 0=-0.8t^2+10t+2 equation:

0=-0.8t^2+10t+2

We move all terms to the left:

0-(-0.8t^2+10t+2)=0

We add all the numbers together, and all the variables

-(-0.8t^2+10t+2)=0

We get rid of parentheses

0.8t^2-10t-2=0

a = 0.8; b = -10; c = -2;
Δ = b2-4ac
Δ = -102-4·0.8·(-2)
Δ = 106.4
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)-\sqrt{106.4}}{2*0.8}=\frac{10-\sqrt{106.4}}{1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{106.4}}{2*0.8}=\frac{10+\sqrt{106.4}}{1.6}
$