(0.80-x)(0.20-x)=(2x)^2

Solution for (0.80-x)(0.20-x)=(2x)^2 equation:

(0.80-x)(0.20-x)=(2x)^2

We move all terms to the left:

(0.80-x)(0.20-x)-((2x)^2)=0

determiningTheFunctionDomain
(0.80-x)(0.20-x)-2x^2=0

We add all the numbers together, and all the variables

-2x^2+(-1x+0.8)(-1x+0.2)=0

We multiply parentheses ..

-2x^2+(+x^2-0.2x-0.8x+0.16)=0

We get rid of parentheses

-2x^2+x^2-0.2x-0.8x+0.16=0

We add all the numbers together, and all the variables

-1x^2-1x+0.16=0

a = -1; b = -1; c = +0.16;
Δ = b2-4ac
Δ = -12-4·(-1)·0.16
Δ = 1.64
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{-(-1)-\sqrt{1.64}}{2*-1}=\frac{1-\sqrt{1.64}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{1.64}}{2*-1}=\frac{1+\sqrt{1.64}}{-2}
$