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g^2-4g=7
We move all terms to the left:
g^2-4g-(7)=0
a = 1; b = -4; c = -7;
Δ = b2-4ac
Δ = -42-4·1·(-7)
Δ = 44
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}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{11}}{2*1}=\frac{4-2\sqrt{11}}{2} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{11}}{2*1}=\frac{4+2\sqrt{11}}{2} $
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