(3n+2)n=5(n-4)

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Solution for (3n+2)n=5(n-4) equation: 



(3n+2)n=5(n-4)

We move all terms to the left:

(3n+2)n-(5(n-4))=0

We multiply parentheses

3n^2+2n-(5(n-4))=0



We calculate terms in parentheses: -(5(n-4)), so:

5(n-4)

We multiply parentheses

5n-20

Back to the equation:

-(5n-20)



We get rid of parentheses

3n^2+2n-5n+20=0

We add all the numbers together, and all the variables

3n^2-3n+20=0

a = 3; b = -3; c = +20;
Δ = b2-4ac
Δ = -32-4·3·20
Δ = -231
Delta is less than zero, so there is no solution for the equation

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