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(G)=3G^2+4G-5
We move all terms to the left:
(G)-(3G^2+4G-5)=0
We get rid of parentheses
-3G^2+G-4G+5=0
We add all the numbers together, and all the variables
-3G^2-3G+5=0
a = -3; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-3)·5
Δ = 69
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{-(-3)-\sqrt{69}}{2*-3}=\frac{3-\sqrt{69}}{-6} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{69}}{2*-3}=\frac{3+\sqrt{69}}{-6} $
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