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6n^2+23n+20=0
a = 6; b = 23; c = +20;
Δ = b2-4ac
Δ = 232-4·6·20
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*6}=\frac{-30}{12} =-2+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*6}=\frac{-16}{12} =-1+1/3 $
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