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39j^2-47j=0
a = 39; b = -47; c = 0;
Δ = b2-4ac
Δ = -472-4·39·0
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2209}=47$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-47}{2*39}=\frac{0}{78} =0 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+47}{2*39}=\frac{94}{78} =1+8/39 $
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