6(x+1)(x-1)=13(x-1)+1

Solution for 6(x+1)(x-1)=13(x-1)+1 equation:

6(x+1)(x-1)=13(x-1)+1

We move all terms to the left:

6(x+1)(x-1)-(13(x-1)+1)=0

We use the square of the difference formula

x^2-(13(x-1)+1)-1=0


We calculate terms in parentheses: -(13(x-1)+1), so:

13(x-1)+1

We multiply parentheses

13x-13+1

We add all the numbers together, and all the variables

13x-12

Back to the equation:

-(13x-12)


We get rid of parentheses

x^2-13x+12-1=0

We add all the numbers together, and all the variables

x^2-13x+11=0

a = 1; b = -13; c = +11;
Δ = b2-4ac
Δ = -132-4·1·11
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5\sqrt{5}}{2*1}=\frac{13-5\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5\sqrt{5}}{2*1}=\frac{13+5\sqrt{5}}{2}
$