(3n+3)(3n+1)=9

Solution for (3n+3)(3n+1)=9 equation:

(3n+3)(3n+1)=9

We move all terms to the left:

(3n+3)(3n+1)-(9)=0

We multiply parentheses ..

(+9n^2+3n+9n+3)-9=0

We get rid of parentheses

9n^2+3n+9n+3-9=0

We add all the numbers together, and all the variables

9n^2+12n-6=0

a = 9; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·9·(-6)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{10}}{2*9}=\frac{-12-6\sqrt{10}}{18}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{10}}{2*9}=\frac{-12+6\sqrt{10}}{18}
$