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(G)=2G^2+3G-3
We move all terms to the left:
(G)-(2G^2+3G-3)=0
We get rid of parentheses
-2G^2+G-3G+3=0
We add all the numbers together, and all the variables
-2G^2-2G+3=0
a = -2; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-2)·3
Δ = 28
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}$$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-2}=\frac{2-2\sqrt{7}}{-4} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-2}=\frac{2+2\sqrt{7}}{-4} $
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