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5/3g=2g-2/3g+2
We move all terms to the left:
5/3g-(2g-2/3g+2)=0
Domain of the equation: 3g!=0
g!=0/3
g!=0
g∈R
Domain of the equation: 3g+2)!=0We get rid of parentheses
g∈R
5/3g-2g+2/3g-2=0
We multiply all the terms by the denominator
-2g*3g-2*3g+5+2=0
We add all the numbers together, and all the variables
-2g*3g-2*3g+7=0
Wy multiply elements
-6g^2-6g+7=0
a = -6; b = -6; c = +7;
Δ = b2-4ac
Δ = -62-4·(-6)·7
Δ = 204
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{51}}{2*-6}=\frac{6-2\sqrt{51}}{-12} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{51}}{2*-6}=\frac{6+2\sqrt{51}}{-12} $
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