x(x-4)=3(x-4)+x

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

x(x-4)=3(x-4)+x

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(3(x-4)+x), so:

3(x-4)+x

We add all the numbers together, and all the variables

x+3(x-4)

We multiply parentheses

x+3x-12

We add all the numbers together, and all the variables

4x-12

Back to the equation:

-(4x-12)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-8x+12=0

a = 1; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·1·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*1}=\frac{12}{2}
=6
$