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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

x^2+(+x^2-2x-3x+6)-3x-((x-2))=0


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

(x-2)

We get rid of parentheses

x-2

Back to the equation:

-(x-2)


We get rid of parentheses

x^2+x^2-2x-3x-3x-x+6+2=0

We add all the numbers together, and all the variables

2x^2-9x+8=0

a = 2; b = -9; c = +8;
Δ = b2-4ac
Δ = -92-4·2·8
Δ = 17
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{-(-9)-\sqrt{17}}{2*2}=\frac{9-\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{17}}{2*2}=\frac{9+\sqrt{17}}{4}
$