3(n^2+3)=11n-6+n^2

Solution for 3(n^2+3)=11n-6+n^2 equation:

3(n^2+3)=11n-6+n^2

We move all terms to the left:

3(n^2+3)-(11n-6+n^2)=0

We multiply parentheses

-(11n-6+n^2)+3n^2+9=0

We get rid of parentheses

-n^2+3n^2-11n+6+9=0

We add all the numbers together, and all the variables

2n^2-11n+15=0

a = 2; b = -11; c = +15;
Δ = b2-4ac
Δ = -112-4·2·15
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2*2}=\frac{10}{4}
=2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2*2}=\frac{12}{4}
=3
$