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6n^2-19n-11=0
a = 6; b = -19; c = -11;
Δ = b2-4ac
Δ = -192-4·6·(-11)
Δ = 625
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{625}=25$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-25}{2*6}=\frac{-6}{12} =-1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+25}{2*6}=\frac{44}{12} =3+2/3 $
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