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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

n(n-13)+16

We multiply parentheses

n^2-13n+16

Back to the equation:

-(n^2-13n+16)


We add all the numbers together, and all the variables

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

We get rid of parentheses

2n^2-n^2+4n+13n-16=0

We add all the numbers together, and all the variables

n^2+17n-16=0

a = 1; b = 17; c = -16;
Δ = b2-4ac
Δ = 172-4·1·(-16)
Δ = 353
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{-(17)-\sqrt{353}}{2*1}=\frac{-17-\sqrt{353}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{353}}{2*1}=\frac{-17+\sqrt{353}}{2}
$