(1-x)(3-x)=(x-2)2

Solution for (1-x)(3-x)=(x-2)2 equation:

(1-x)(3-x)=(x-2)2

We move all terms to the left:

(1-x)(3-x)-((x-2)2)=0

We add all the numbers together, and all the variables

(-1x+1)(-1x+3)-((x-2)2)=0

We multiply parentheses ..

(+x^2-3x-1x+3)-((x-2)2)=0


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

(x-2)2

We multiply parentheses

2x-4

Back to the equation:

-(2x-4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-6x+7=0

a = 1; b = -6; c = +7;
Δ = b2-4ac
Δ = -62-4·1·7
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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