-6(n-3)n=5(4-n)

Solution for -6(n-3)n=5(4-n) equation:

-6(n-3)n=5(4-n)

We move all terms to the left:

-6(n-3)n-(5(4-n))=0

We add all the numbers together, and all the variables

-6(n-3)n-(5(-1n+4))=0

We multiply parentheses

-6n^2+18n-(5(-1n+4))=0


We calculate terms in parentheses: -(5(-1n+4)), so:

5(-1n+4)

We multiply parentheses

-5n+20

Back to the equation:

-(-5n+20)


We get rid of parentheses

-6n^2+18n+5n-20=0

We add all the numbers together, and all the variables

-6n^2+23n-20=0

a = -6; b = 23; c = -20;
Δ = b2-4ac
Δ = 232-4·(-6)·(-20)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*-6}=\frac{-30}{-12}
=2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*-6}=\frac{-16}{-12}
=1+1/3
$