28=18n-36+3n^2-12n+12=3n^2+6n+4

Solution for 28=18n-36+3n^2-12n+12=3n^2+6n+4 equation:

28=18n-36+3n^2-12n+12=3n^2+6n+4

We move all terms to the left:

28-(18n-36+3n^2-12n+12)=0

We get rid of parentheses

-3n^2-18n+12n+36-12+28=0

We add all the numbers together, and all the variables

-3n^2-6n+52=0

a = -3; b = -6; c = +52;
Δ = b2-4ac
Δ = -62-4·(-3)·52
Δ = 660
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:

$\sqrt{\Delta}=\sqrt{660}=\sqrt{4*165}=\sqrt{4}*\sqrt{165}=2\sqrt{165}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{165}}{2*-3}=\frac{6-2\sqrt{165}}{-6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{165}}{2*-3}=\frac{6+2\sqrt{165}}{-6}
$