3n(n+2)=9(6-n)

Solution for 3n(n+2)=9(6-n) equation:

3n(n+2)=9(6-n)

We move all terms to the left:

3n(n+2)-(9(6-n))=0

We add all the numbers together, and all the variables

3n(n+2)-(9(-1n+6))=0

We multiply parentheses

3n^2+6n-(9(-1n+6))=0


We calculate terms in parentheses: -(9(-1n+6)), so:

9(-1n+6)

We multiply parentheses

-9n+54

Back to the equation:

-(-9n+54)


We get rid of parentheses

3n^2+6n+9n-54=0

We add all the numbers together, and all the variables

3n^2+15n-54=0

a = 3; b = 15; c = -54;
Δ = b2-4ac
Δ = 152-4·3·(-54)
Δ = 873
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:

$\sqrt{\Delta}=\sqrt{873}=\sqrt{9*97}=\sqrt{9}*\sqrt{97}=3\sqrt{97}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{97}}{2*3}=\frac{-15-3\sqrt{97}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{97}}{2*3}=\frac{-15+3\sqrt{97}}{6}
$