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65n^2+20n=0
a = 65; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·65·0
Δ = 400
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{400}=20$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*65}=\frac{-40}{130} =-4/13 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*65}=\frac{0}{130} =0 $
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