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32n^2-24n=0
a = 32; b = -24; c = 0;
Δ = b2-4ac
Δ = -242-4·32·0
Δ = 576
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{576}=24$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24}{2*32}=\frac{0}{64} =0 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24}{2*32}=\frac{48}{64} =3/4 $
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