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(G)=1/2G+5
We move all terms to the left:
(G)-(1/2G+5)=0
Domain of the equation: 2G+5)!=0We get rid of parentheses
G∈R
G-1/2G-5=0
We multiply all the terms by the denominator
G*2G-5*2G-1=0
Wy multiply elements
2G^2-10G-1=0
a = 2; b = -10; c = -1;
Δ = b2-4ac
Δ = -102-4·2·(-1)
Δ = 108
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}$$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6\sqrt{3}}{2*2}=\frac{10-6\sqrt{3}}{4} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6\sqrt{3}}{2*2}=\frac{10+6\sqrt{3}}{4} $
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