n(n-1)+n(n+8)=n(n-13)+n(n+1)16

Solution for n(n-1)+n(n+8)=n(n-13)+n(n+1)16 equation:

n(n-1)+n(n+8)=n(n-13)+n(n+1)16

We move all terms to the left:

n(n-1)+n(n+8)-(n(n-13)+n(n+1)16)=0

We multiply parentheses

n^2+n^2-1n+8n-(n(n-13)+n(n+1)16)=0


We calculate terms in parentheses: -(n(n-13)+n(n+1)16), so:

n(n-13)+n(n+1)16

We multiply parentheses

n^2+16n^2-13n+16n

We add all the numbers together, and all the variables

17n^2+3n

Back to the equation:

-(17n^2+3n)


We add all the numbers together, and all the variables

2n^2+7n-(17n^2+3n)=0

We get rid of parentheses

2n^2-17n^2+7n-3n=0

We add all the numbers together, and all the variables

-15n^2+4n=0

a = -15; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-15)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-15}=\frac{-8}{-30}
=4/15
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-15}=\frac{0}{-30}
=0
$