(x-1)(x+1)=(x+1)(3x-5)

Solution for (x-1)(x+1)=(x+1)(3x-5) equation:

(x-1)(x+1)=(x+1)(3x-5)

We move all terms to the left:

(x-1)(x+1)-((x+1)(3x-5))=0

We use the square of the difference formula

x^2-((x+1)(3x-5))-1=0

We multiply parentheses ..

x^2-((+3x^2-5x+3x-5))-1=0


We calculate terms in parentheses: -((+3x^2-5x+3x-5)), so:

(+3x^2-5x+3x-5)

We get rid of parentheses

3x^2-5x+3x-5

We add all the numbers together, and all the variables

3x^2-2x-5

Back to the equation:

-(3x^2-2x-5)


We get rid of parentheses

x^2-3x^2+2x+5-1=0

We add all the numbers together, and all the variables

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

a = -2; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-2)·4
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*-2}=\frac{4}{-4}
=-1
$