(2-x)(1-x)+0.2=0

Solution for (2-x)(1-x)+0.2=0 equation:

(2-x)(1-x)+0.2=0

We add all the numbers together, and all the variables

(-1x+2)(-1x+1)+0.2=0

We multiply parentheses ..

(+x^2-1x-2x+2)+0.2=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-3x+2.2=0

a = 1; b = -3; c = +2.2;
Δ = b2-4ac
Δ = -32-4·1·2.2
Δ = 0.2
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{-(-3)-\sqrt{0.2}}{2*1}=\frac{3-\sqrt{0.2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{0.2}}{2*1}=\frac{3+\sqrt{0.2}}{2}
$