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(G)=3/5G-9
We move all terms to the left:
(G)-(3/5G-9)=0
Domain of the equation: 5G-9)!=0We get rid of parentheses
G∈R
G-3/5G+9=0
We multiply all the terms by the denominator
G*5G+9*5G-3=0
Wy multiply elements
5G^2+45G-3=0
a = 5; b = 45; c = -3;
Δ = b2-4ac
Δ = 452-4·5·(-3)
Δ = 2085
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{-(45)-\sqrt{2085}}{2*5}=\frac{-45-\sqrt{2085}}{10} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{2085}}{2*5}=\frac{-45+\sqrt{2085}}{10} $
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