(x-4)(3x-2)=(x-4)

Solution for (x-4)(3x-2)=(x-4) equation:

(x-4)(3x-2)=(x-4)

We move all terms to the left:

(x-4)(3x-2)-((x-4))=0

We multiply parentheses ..

(+3x^2-2x-12x+8)-((x-4))=0


We calculate terms in parentheses: -((x-4)), so:

(x-4)

We get rid of parentheses

x-4

Back to the equation:

-(x-4)


We get rid of parentheses

3x^2-2x-12x-x+8+4=0

We add all the numbers together, and all the variables

3x^2-15x+12=0

a = 3; b = -15; c = +12;
Δ = b2-4ac
Δ = -152-4·3·12
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-9}{2*3}=\frac{6}{6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+9}{2*3}=\frac{24}{6}
=4
$