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18n^2-39n+20=0
a = 18; b = -39; c = +20;
Δ = b2-4ac
Δ = -392-4·18·20
Δ = 81
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{81}=9$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-9}{2*18}=\frac{30}{36} =5/6 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+9}{2*18}=\frac{48}{36} =1+1/3 $
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