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2x+3/3=20/3x
We move all terms to the left:
2x+3/3-(20/3x)=0
Domain of the equation: 3x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
2x-(+20/3x)+3/3=0
We add all the numbers together, and all the variables
2x-(+20/3x)+1=0
We get rid of parentheses
2x-20/3x+1=0
We multiply all the terms by the denominator
2x*3x+1*3x-20=0
Wy multiply elements
6x^2+3x-20=0
a = 6; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·6·(-20)
Δ = 489
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{489}}{2*6}=\frac{-3-\sqrt{489}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{489}}{2*6}=\frac{-3+\sqrt{489}}{12} $
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