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0.3n^2-2n=0
a = 0.3; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·0.3·0
Δ = 4
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{4}=2$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*0.3}=\frac{0}{0.6} =0 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*0.3}=\frac{4}{0.6} =6+0.4/0.6 $
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