(x-4)-(x+1)=(x-4)(x+2)

Solution for (x-4)-(x+1)=(x-4)(x+2) equation:

(x-4)-(x+1)=(x-4)(x+2)

We move all terms to the left:

(x-4)-(x+1)-((x-4)(x+2))=0

We get rid of parentheses

x-x-((x-4)(x+2))-4-1=0

We multiply parentheses ..

-((+x^2+2x-4x-8))+x-x-4-1=0


We calculate terms in parentheses: -((+x^2+2x-4x-8)), so:

(+x^2+2x-4x-8)

We get rid of parentheses

x^2+2x-4x-8

We add all the numbers together, and all the variables

x^2-2x-8

Back to the equation:

-(x^2-2x-8)


We add all the numbers together, and all the variables

-(x^2-2x-8)-5=0

We get rid of parentheses

-x^2+2x+8-5=0

We add all the numbers together, and all the variables

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

a = -1; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-1)·3
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{-(2)-4}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*-1}=\frac{2}{-2}
=-1
$