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(G)=-3G^2-12G-9
We move all terms to the left:
(G)-(-3G^2-12G-9)=0
We get rid of parentheses
3G^2+12G+G+9=0
We add all the numbers together, and all the variables
3G^2+13G+9=0
a = 3; b = 13; c = +9;
Δ = b2-4ac
Δ = 132-4·3·9
Δ = 61
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{-(13)-\sqrt{61}}{2*3}=\frac{-13-\sqrt{61}}{6} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{61}}{2*3}=\frac{-13+\sqrt{61}}{6} $
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