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(G)=17-2/3G
We move all terms to the left:
(G)-(17-2/3G)=0
Domain of the equation: 3G)!=0We add all the numbers together, and all the variables
G!=0/1
G!=0
G∈R
G-(-2/3G+17)=0
We get rid of parentheses
G+2/3G-17=0
We multiply all the terms by the denominator
G*3G-17*3G+2=0
Wy multiply elements
3G^2-51G+2=0
a = 3; b = -51; c = +2;
Δ = b2-4ac
Δ = -512-4·3·2
Δ = 2577
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}$$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-51)-\sqrt{2577}}{2*3}=\frac{51-\sqrt{2577}}{6} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-51)+\sqrt{2577}}{2*3}=\frac{51+\sqrt{2577}}{6} $
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