0=x^2+0.2x-0.02

Solution for 0=x^2+0.2x-0.02 equation:

0=x^2+0.2x-0.02

We move all terms to the left:

0-(x^2+0.2x-0.02)=0

We add all the numbers together, and all the variables

-(x^2+0.2x-0.02)=0

We get rid of parentheses

-x^2-0.2x+0.02=0

We add all the numbers together, and all the variables

-1x^2-0.2x+0.02=0

a = -1; b = -0.2; c = +0.02;
Δ = b2-4ac
Δ = -0.22-4·(-1)·0.02
Δ = 0.12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.2)-\sqrt{0.12}}{2*-1}=\frac{0.2-\sqrt{0.12}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.2)+\sqrt{0.12}}{2*-1}=\frac{0.2+\sqrt{0.12}}{-2}
$