2(n-4)=3(n+4)2n-8=3n+12

Solution for 2(n-4)=3(n+4)2n-8=3n+12 equation:

2(n-4)=3(n+4)2n-8=3n+12

We move all terms to the left:

2(n-4)-(3(n+4)2n-8)=0

We multiply parentheses

2n-(3(n+4)2n-8)-8=0


We calculate terms in parentheses: -(3(n+4)2n-8), so:

3(n+4)2n-8

We multiply parentheses

6n^2+24n-8

Back to the equation:

-(6n^2+24n-8)


We get rid of parentheses

-6n^2+2n-24n+8-8=0

We add all the numbers together, and all the variables

-6n^2-22n=0

a = -6; b = -22; c = 0;
Δ = b2-4ac
Δ = -222-4·(-6)·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-22}{2*-6}=\frac{0}{-12}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+22}{2*-6}=\frac{44}{-12}
=-3+2/3
$