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(G)=5/2G
We move all terms to the left:
(G)-(5/2G)=0
Domain of the equation: 2G)!=0We add all the numbers together, and all the variables
G!=0/1
G!=0
G∈R
G-(+5/2G)=0
We get rid of parentheses
G-5/2G=0
We multiply all the terms by the denominator
G*2G-5=0
Wy multiply elements
2G^2-5=0
a = 2; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·2·(-5)
Δ = 40
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}$$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*2}=\frac{0-2\sqrt{10}}{4} =-\frac{2\sqrt{10}}{4} =-\frac{\sqrt{10}}{2} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*2}=\frac{0+2\sqrt{10}}{4} =\frac{2\sqrt{10}}{4} =\frac{\sqrt{10}}{2} $
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