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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2-3x-((x-4))-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

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

We add all the numbers together, and all the variables

x^2-4x=0

a = 1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·1·0
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{-(-4)-4}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*1}=\frac{8}{2}
=4
$