-3(n+4)2n=-2(n+6)

Solution for -3(n+4)2n=-2(n+6) equation:

-3(n+4)2n=-2(n+6)

We move all terms to the left:

-3(n+4)2n-(-2(n+6))=0

We multiply parentheses

-6n^2-24n-(-2(n+6))=0


We calculate terms in parentheses: -(-2(n+6)), so:

-2(n+6)

We multiply parentheses

-2n-12

Back to the equation:

-(-2n-12)


We get rid of parentheses

-6n^2-24n+2n+12=0

We add all the numbers together, and all the variables

-6n^2-22n+12=0

a = -6; b = -22; c = +12;
Δ = b2-4ac
Δ = -222-4·(-6)·12
Δ = 772
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{193}}{2*-6}=\frac{22-2\sqrt{193}}{-12}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{193}}{2*-6}=\frac{22+2\sqrt{193}}{-12}
$