2n+14=n(n+1)

Solution for 2n+14=n(n+1) equation:

2n+14=n(n+1)

We move all terms to the left:

2n+14-(n(n+1))=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+14=0

We add all the numbers together, and all the variables

-1n^2+n+14=0

a = -1; b = 1; c = +14;
Δ = b2-4ac
Δ = 12-4·(-1)·14
Δ = 57
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{-(1)-\sqrt{57}}{2*-1}=\frac{-1-\sqrt{57}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{57}}{2*-1}=\frac{-1+\sqrt{57}}{-2}
$