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(G)=G2+15G-54
We move all terms to the left:
(G)-(G2+15G-54)=0
We add all the numbers together, and all the variables
-(+G^2+15G-54)+G=0
We get rid of parentheses
-G^2-15G+G+54=0
We add all the numbers together, and all the variables
-1G^2-14G+54=0
a = -1; b = -14; c = +54;
Δ = b2-4ac
Δ = -142-4·(-1)·54
Δ = 412
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}$$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{103}}{2*-1}=\frac{14-2\sqrt{103}}{-2} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{103}}{2*-1}=\frac{14+2\sqrt{103}}{-2} $
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