(x+3)(x-1)-(x+1)(x-1)=+8

Solution for (x+3)(x-1)-(x+1)(x-1)=+8 equation:

(x+3)(x-1)-(x+1)(x-1)=+8

We move all terms to the left:

(x+3)(x-1)-(x+1)(x-1)-(+8)=0

We add all the numbers together, and all the variables

(x+3)(x-1)-(x+1)(x-1)-8=0

We use the square of the difference formula

x^2+(x+3)(x-1)+1-8=0

We multiply parentheses ..

x^2+(+x^2-1x+3x-3)+1-8=0

We add all the numbers together, and all the variables

x^2+(+x^2-1x+3x-3)-7=0

We get rid of parentheses

x^2+x^2-1x+3x-3-7=0

We add all the numbers together, and all the variables

2x^2+2x-10=0

a = 2; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·2·(-10)
Δ = 84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{21}}{2*2}=\frac{-2-2\sqrt{21}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{21}}{2*2}=\frac{-2+2\sqrt{21}}{4}
$