0=0.003x^2+0.2x-2

Solution for 0=0.003x^2+0.2x-2 equation:

0=0.003x^2+0.2x-2

We move all terms to the left:

0-(0.003x^2+0.2x-2)=0

We add all the numbers together, and all the variables

-(0.003x^2+0.2x-2)=0

We get rid of parentheses

-0.003x^2-0.2x+2=0

a = -0.003; b = -0.2; c = +2;
Δ = b2-4ac
Δ = -0.22-4·(-0.003)·2
Δ = 0.064
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{0.064}}{2*-0.003}=\frac{0.2-\sqrt{0.064}}{-0.006}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.2)+\sqrt{0.064}}{2*-0.003}=\frac{0.2+\sqrt{0.064}}{-0.006}
$