n(n+3)=8(n+3)

Solution for n(n+3)=8(n+3) equation:

n(n+3)=8(n+3)

We move all terms to the left:

n(n+3)-(8(n+3))=0

We multiply parentheses

n^2+3n-(8(n+3))=0


We calculate terms in parentheses: -(8(n+3)), so:

8(n+3)

We multiply parentheses

8n+24

Back to the equation:

-(8n+24)


We get rid of parentheses

n^2+3n-8n-24=0

We add all the numbers together, and all the variables

n^2-5n-24=0

a = 1; b = -5; c = -24;
Δ = b2-4ac
Δ = -52-4·1·(-24)
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{-(-5)-11}{2*1}=\frac{-6}{2}
=-3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*1}=\frac{16}{2}
=8
$