2+4+6+8+10+2n=n(n+1)

Solution for 2+4+6+8+10+2n=n(n+1) equation:

2+4+6+8+10+2n=n(n+1)

We move all terms to the left:

2+4+6+8+10+2n-(n(n+1))=0

We add all the numbers together, and all the variables

2n-(n(n+1))+30=0


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

n(n+1)

We multiply parentheses

n^2+n

Back to the equation:

-(n^2+n)


We get rid of parentheses

-n^2+2n-n+30=0

We add all the numbers together, and all the variables

-1n^2+n+30=0

a = -1; b = 1; c = +30;
Δ = b2-4ac
Δ = 12-4·(-1)·30
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*-1}=\frac{-12}{-2}
=+6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*-1}=\frac{10}{-2}
=-5
$