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6n^2-19n+15=0
a = 6; b = -19; c = +15;
Δ = b2-4ac
Δ = -192-4·6·15
Δ = 1
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{1}=1$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*6}=\frac{18}{12} =1+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*6}=\frac{20}{12} =1+2/3 $
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