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2v+2/3v=4
We move all terms to the left:
2v+2/3v-(4)=0
Domain of the equation: 3v!=0We multiply all the terms by the denominator
v!=0/3
v!=0
v∈R
2v*3v-4*3v+2=0
Wy multiply elements
6v^2-12v+2=0
a = 6; b = -12; c = +2;
Δ = b2-4ac
Δ = -122-4·6·2
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{6}}{2*6}=\frac{12-4\sqrt{6}}{12} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{6}}{2*6}=\frac{12+4\sqrt{6}}{12} $
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