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2x+1/3x-2=0
Domain of the equation: 3x!=0We multiply all the terms by the denominator
x!=0/3
x!=0
x∈R
2x*3x-2*3x+1=0
Wy multiply elements
6x^2-6x+1=0
a = 6; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·6·1
Δ = 12
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{3}}{2*6}=\frac{6-2\sqrt{3}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{3}}{2*6}=\frac{6+2\sqrt{3}}{12} $
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