x2-4x+4=(1-x)(x-2)

Solution for x2-4x+4=(1-x)(x-2) equation:

x2-4x+4=(1-x)(x-2)

We move all terms to the left:

x2-4x+4-((1-x)(x-2))=0

We add all the numbers together, and all the variables

x2-4x-((-1x+1)(x-2))+4=0

We add all the numbers together, and all the variables

x^2-4x-((-1x+1)(x-2))+4=0

We multiply parentheses ..

x^2-((-1x^2+2x+x-2))-4x+4=0


We calculate terms in parentheses: -((-1x^2+2x+x-2)), so:

(-1x^2+2x+x-2)

We get rid of parentheses

-1x^2+2x+x-2

We add all the numbers together, and all the variables

-1x^2+3x-2

Back to the equation:

-(-1x^2+3x-2)


We get rid of parentheses

x^2+1x^2-3x-4x+2+4=0

We add all the numbers together, and all the variables

2x^2-7x+6=0

a = 2; b = -7; c = +6;
Δ = b2-4ac
Δ = -72-4·2·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*2}=\frac{6}{4}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*2}=\frac{8}{4}
=2
$