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(G)=13G^2-48
We move all terms to the left:
(G)-(13G^2-48)=0
We get rid of parentheses
-13G^2+G+48=0
a = -13; b = 1; c = +48;
Δ = b2-4ac
Δ = 12-4·(-13)·48
Δ = 2497
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{-(1)-\sqrt{2497}}{2*-13}=\frac{-1-\sqrt{2497}}{-26} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{2497}}{2*-13}=\frac{-1+\sqrt{2497}}{-26} $
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