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^3-N^2-12N=0
We add all the numbers together, and all the variables
-1N^2-12N=0
a = -1; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·(-1)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*-1}=\frac{0}{-2} =0 $$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*-1}=\frac{24}{-2} =-12 $
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