(3n-1)(3n+1)=12n+8

Solution for (3n-1)(3n+1)=12n+8 equation:

(3n-1)(3n+1)=12n+8

We move all terms to the left:

(3n-1)(3n+1)-(12n+8)=0

We use the square of the difference formula

9n^2-(12n+8)-1=0

We get rid of parentheses

9n^2-12n-8-1=0

We add all the numbers together, and all the variables

9n^2-12n-9=0

a = 9; b = -12; c = -9;
Δ = b2-4ac
Δ = -122-4·9·(-9)
Δ = 468
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{13}}{2*9}=\frac{12-6\sqrt{13}}{18}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{13}}{2*9}=\frac{12+6\sqrt{13}}{18}
$