6(2u+3)=3(2/3u+1)

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Solution for 6(2u+3)=3(2/3u+1) equation:



6(2u+3)=3(2/3u+1)
We move all terms to the left:
6(2u+3)-(3(2/3u+1))=0
Domain of the equation: 3u+1))!=0
u∈R
We multiply parentheses
12u-(3(2/3u+1))+18=0
We multiply all the terms by the denominator
12u*3u+18*3u+1))-(3(2+1))=0
We add all the numbers together, and all the variables
12u*3u+18*3u+1))-(33)=0
We add all the numbers together, and all the variables
12u*3u+18*3u=0
Wy multiply elements
36u^2+54u=0
a = 36; b = 54; c = 0;
Δ = b2-4ac
Δ = 542-4·36·0
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-54}{2*36}=\frac{-108}{72} =-1+1/2 $
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+54}{2*36}=\frac{0}{72} =0 $

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