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18j^2+41j=0
a = 18; b = 41; c = 0;
Δ = b2-4ac
Δ = 412-4·18·0
Δ = 1681
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{1681}=41$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-41}{2*18}=\frac{-82}{36} =-2+5/18 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+41}{2*18}=\frac{0}{36} =0 $
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