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32n^2-56n+24=0
a = 32; b = -56; c = +24;
Δ = b2-4ac
Δ = -562-4·32·24
Δ = 64
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{64}=8$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-8}{2*32}=\frac{48}{64} =3/4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+8}{2*32}=\frac{64}{64} =1 $
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