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47g^2-13g=0
a = 47; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·47·0
Δ = 169
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}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*47}=\frac{0}{94} =0 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*47}=\frac{26}{94} =13/47 $
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