0=-5t^2+2t+0.2

Solution for 0=-5t^2+2t+0.2 equation:

0=-5t^2+2t+0.2

We move all terms to the left:

0-(-5t^2+2t+0.2)=0

We add all the numbers together, and all the variables

-(-5t^2+2t+0.2)=0

We get rid of parentheses

5t^2-2t-0.2=0

a = 5; b = -2; c = -0.2;
Δ = b2-4ac
Δ = -22-4·5·(-0.2)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{2}}{2*5}=\frac{2-2\sqrt{2}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{2}}{2*5}=\frac{2+2\sqrt{2}}{10}
$