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20n^2+23n+6=0
a = 20; b = 23; c = +6;
Δ = b2-4ac
Δ = 232-4·20·6
Δ = 49
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{49}=7$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*20}=\frac{-30}{40} =-3/4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*20}=\frac{-16}{40} =-2/5 $
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