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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(9n(n-3)), so:

9n(n-3)

We multiply parentheses

9n^2-27n

Back to the equation:

-(9n^2-27n)


We get rid of parentheses

-9n^2+6n+27n+9=0

We add all the numbers together, and all the variables

-9n^2+33n+9=0

a = -9; b = 33; c = +9;
Δ = b2-4ac
Δ = 332-4·(-9)·9
Δ = 1413
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:

$\sqrt{\Delta}=\sqrt{1413}=\sqrt{9*157}=\sqrt{9}*\sqrt{157}=3\sqrt{157}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{157}}{2*-9}=\frac{-33-3\sqrt{157}}{-18}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{157}}{2*-9}=\frac{-33+3\sqrt{157}}{-18}
$