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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(5(n-4)*2n), so:

5(n-4)*2n

We multiply parentheses

10n^2-40n

Back to the equation:

-(10n^2-40n)


We get rid of parentheses

3n^2-10n^2+2n+40n=0

We add all the numbers together, and all the variables

-7n^2+42n=0

a = -7; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·(-7)·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*-7}=\frac{-84}{-14}
=+6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*-7}=\frac{0}{-14}
=0
$