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122/3x+41/6=4051/2x
We move all terms to the left:
122/3x+41/6-(4051/2x)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 2x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
122/3x-(+4051/2x)+41/6=0
We get rid of parentheses
122/3x-4051/2x+41/6=0
We calculate fractions
492x^2/216x^2+8784x/216x^2+(-437508x)/216x^2=0
We multiply all the terms by the denominator
492x^2+8784x+(-437508x)=0
We get rid of parentheses
492x^2+8784x-437508x=0
We add all the numbers together, and all the variables
492x^2-428724x=0
a = 492; b = -428724; c = 0;
Δ = b2-4ac
Δ = -4287242-4·492·0
Δ = 183804268176
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{183804268176}=428724$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-428724)-428724}{2*492}=\frac{0}{984} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-428724)+428724}{2*492}=\frac{857448}{984} =871+16/41 $
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