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(G)=G^2-5G+6
We move all terms to the left:
(G)-(G^2-5G+6)=0
We get rid of parentheses
-G^2+G+5G-6=0
We add all the numbers together, and all the variables
-1G^2+6G-6=0
a = -1; b = 6; c = -6;
Δ = b2-4ac
Δ = 62-4·(-1)·(-6)
Δ = 12
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}$$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3}}{2*-1}=\frac{-6-2\sqrt{3}}{-2} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3}}{2*-1}=\frac{-6+2\sqrt{3}}{-2} $
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