3(3+3n)n=4

Solution for 3(3+3n)n=4 equation:

3(3+3n)n=4

We move all terms to the left:

3(3+3n)n-(4)=0

We add all the numbers together, and all the variables

3(3n+3)n-4=0

We multiply parentheses

9n^2+9n-4=0

a = 9; b = 9; c = -4;
Δ = b2-4ac
Δ = 92-4·9·(-4)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*9}=\frac{-24}{18}
=-1+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*9}=\frac{6}{18}
=1/3
$