(0.2+x)(0.1+x)=0

Solution for (0.2+x)(0.1+x)=0 equation:

(0.2+x)(0.1+x)=0

We add all the numbers together, and all the variables

(x+0.2)(x+0.1)=0

We multiply parentheses ..

(+x^2+0.1x+0.2x+0.02)=0

We get rid of parentheses

x^2+0.1x+0.2x+0.02=0

We add all the numbers together, and all the variables

x^2+0.3x+0.02=0

a = 1; b = 0.3; c = +0.02;
Δ = b2-4ac
Δ = 0.32-4·1·0.02
Δ = 0.01
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{-(0.3)-\sqrt{0.01}}{2*1}=\frac{-0.3-\sqrt{0.01}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{0.01}}{2*1}=\frac{-0.3+\sqrt{0.01}}{2}
$