(x+1)(x-1)(x^2+7x)(x-4)-2=2x

Solution for (x+1)(x-1)(x^2+7x)(x-4)-2=2x equation:

(x+1)(x-1)(x^2+7x)(x-4)-2=2x

We move all terms to the left:

(x+1)(x-1)(x^2+7x)(x-4)-2-(2x)=0

We add all the numbers together, and all the variables

-2x+(x+1)(x-1)(x^2+7x)(x-4)-2=0

We use the square of the difference formula

x^2-2x-1-2=0

We add all the numbers together, and all the variables

x^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{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*1}=\frac{6}{2}
=3
$