2n(n+4)+18=n(n+5)+n(n+5)+n(n-2)-7

Solution for 2n(n+4)+18=n(n+5)+n(n+5)+n(n-2)-7 equation:

2n(n+4)+18=n(n+5)+n(n+5)+n(n-2)-7

We move all terms to the left:

2n(n+4)+18-(n(n+5)+n(n+5)+n(n-2)-7)=0

We multiply parentheses

2n^2+8n-(n(n+5)+n(n+5)+n(n-2)-7)+18=0


We calculate terms in parentheses: -(n(n+5)+n(n+5)+n(n-2)-7), so:

n(n+5)+n(n+5)+n(n-2)-7

We multiply parentheses

n^2+n^2+n^2+5n+5n-2n-7

We add all the numbers together, and all the variables

3n^2+8n-7

Back to the equation:

-(3n^2+8n-7)


We get rid of parentheses

2n^2-3n^2+8n-8n+7+18=0

We add all the numbers together, and all the variables

-1n^2+25=0

a = -1; b = 0; c = +25;
Δ = b2-4ac
Δ = 02-4·(-1)·25
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*-1}=\frac{-10}{-2}
=+5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*-1}=\frac{10}{-2}
=-5
$