n(n-4)+n(n+8)=n(n-13)+n(n-13)+n(n+1)+16

Solution for n(n-4)+n(n+8)=n(n-13)+n(n-13)+n(n+1)+16 equation:

n(n-4)+n(n+8)=n(n-13)+n(n-13)+n(n+1)+16

We move all terms to the left:

n(n-4)+n(n+8)-(n(n-13)+n(n-13)+n(n+1)+16)=0

We multiply parentheses

n^2+n^2-4n+8n-(n(n-13)+n(n-13)+n(n+1)+16)=0


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

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

We multiply parentheses

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

We add all the numbers together, and all the variables

3n^2-25n+16

Back to the equation:

-(3n^2-25n+16)


We add all the numbers together, and all the variables

2n^2+4n-(3n^2-25n+16)=0

We get rid of parentheses

2n^2-3n^2+4n+25n-16=0

We add all the numbers together, and all the variables

-1n^2+29n-16=0

a = -1; b = 29; c = -16;
Δ = b2-4ac
Δ = 292-4·(-1)·(-16)
Δ = 777
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{-(29)-\sqrt{777}}{2*-1}=\frac{-29-\sqrt{777}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{777}}{2*-1}=\frac{-29+\sqrt{777}}{-2}
$