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

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
$