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42j^2+11j=0
a = 42; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·42·0
Δ = 121
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{121}=11$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*42}=\frac{-22}{84} =-11/42 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*42}=\frac{0}{84} =0 $
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