3n^2+n+3=(n+1)(n+2)+1

Solution for 3n^2+n+3=(n+1)(n+2)+1 equation:

3n^2+n+3=(n+1)(n+2)+1

We move all terms to the left:

3n^2+n+3-((n+1)(n+2)+1)=0

We multiply parentheses ..

3n^2-((+n^2+2n+n+2)+1)+n+3=0


We calculate terms in parentheses: -((+n^2+2n+n+2)+1), so:

(+n^2+2n+n+2)+1

We get rid of parentheses

n^2+2n+n+2+1

We add all the numbers together, and all the variables

n^2+3n+3

Back to the equation:

-(n^2+3n+3)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2-2n=0

a = 2; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·2·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*2}=\frac{0}{4}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*2}=\frac{4}{4}
=1
$