-3(1+n)-2n=-6n(n-2)

Solution for -3(1+n)-2n=-6n(n-2) equation:

-3(1+n)-2n=-6n(n-2)

We move all terms to the left:

-3(1+n)-2n-(-6n(n-2))=0

We add all the numbers together, and all the variables

-3(n+1)-2n-(-6n(n-2))=0

We add all the numbers together, and all the variables

-2n-3(n+1)-(-6n(n-2))=0

We multiply parentheses

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


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

-6n(n-2)

We multiply parentheses

-6n^2+12n

Back to the equation:

-(-6n^2+12n)


We add all the numbers together, and all the variables

-(-6n^2+12n)-5n-3=0

We get rid of parentheses

6n^2-12n-5n-3=0

We add all the numbers together, and all the variables

6n^2-17n-3=0

a = 6; b = -17; c = -3;
Δ = b2-4ac
Δ = -172-4·6·(-3)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-19}{2*6}=\frac{-2}{12}
=-1/6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+19}{2*6}=\frac{36}{12}
=3
$