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3-3x=2/3x+6
We move all terms to the left:
3-3x-(2/3x+6)=0
Domain of the equation: 3x+6)!=0We get rid of parentheses
x∈R
-3x-2/3x-6+3=0
We multiply all the terms by the denominator
-3x*3x-6*3x+3*3x-2=0
Wy multiply elements
-9x^2-18x+9x-2=0
We add all the numbers together, and all the variables
-9x^2-9x-2=0
a = -9; b = -9; c = -2;
Δ = b2-4ac
Δ = -92-4·(-9)·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3}{2*-9}=\frac{6}{-18} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3}{2*-9}=\frac{12}{-18} =-2/3 $
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