0=0.2t^2-1.02t

Solution for 0=0.2t^2-1.02t equation:

0=0.2t^2-1.02t

We move all terms to the left:

0-(0.2t^2-1.02t)=0

We add all the numbers together, and all the variables

-(0.2t^2-1.02t)=0

We get rid of parentheses

-0.2t^2+1.02t=0

a = -0.2; b = 1.02; c = 0;
Δ = b2-4ac
Δ = 1.022-4·(-0.2)·0
Δ = 1.0404
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{-(1.02)-\sqrt{1.0404}}{2*-0.2}=\frac{-1.02-\sqrt{1.0404}}{-0.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.02)+\sqrt{1.0404}}{2*-0.2}=\frac{-1.02+\sqrt{1.0404}}{-0.4}
$