12n(n-3)+n+3=396

Solution for 12n(n-3)+n+3=396 equation:

12n(n-3)+n+3=396

We move all terms to the left:

12n(n-3)+n+3-(396)=0

We add all the numbers together, and all the variables

n+12n(n-3)-393=0

We multiply parentheses

12n^2+n-36n-393=0

We add all the numbers together, and all the variables

12n^2-35n-393=0

a = 12; b = -35; c = -393;
Δ = b2-4ac
Δ = -352-4·12·(-393)
Δ = 20089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-\sqrt{20089}}{2*12}=\frac{35-\sqrt{20089}}{24}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{20089}}{2*12}=\frac{35+\sqrt{20089}}{24}
$