2n(n+2)+3n(n-2)+1=17

Solution for 2n(n+2)+3n(n-2)+1=17 equation:

2n(n+2)+3n(n-2)+1=17

We move all terms to the left:

2n(n+2)+3n(n-2)+1-(17)=0

We add all the numbers together, and all the variables

2n(n+2)+3n(n-2)-16=0

We multiply parentheses

2n^2+3n^2+4n-6n-16=0

We add all the numbers together, and all the variables

5n^2-2n-16=0

a = 5; b = -2; c = -16;
Δ = b2-4ac
Δ = -22-4·5·(-16)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-18}{2*5}=\frac{-16}{10}
=-1+3/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+18}{2*5}=\frac{20}{10}
=2
$