3(n+3)=2n(3-1)

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Solution for 3(n+3)=2n(3-1) equation:



3(n+3)=2n(3-1)
We move all terms to the left:
3(n+3)-(2n(3-1))=0
We add all the numbers together, and all the variables
3(n+3)-(2n2)=0
We add all the numbers together, and all the variables
-2n^2+3(n+3)=0
We multiply parentheses
-2n^2+3n+9=0
a = -2; b = 3; c = +9;
Δ = b2-4ac
Δ = 32-4·(-2)·9
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*-2}=\frac{-12}{-4} =+3 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*-2}=\frac{6}{-4} =-1+1/2 $

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