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6n^2-2n-48=0
a = 6; b = -2; c = -48;
Δ = b2-4ac
Δ = -22-4·6·(-48)
Δ = 1156
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{1156}=34$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-34}{2*6}=\frac{-32}{12} =-2+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+34}{2*6}=\frac{36}{12} =3 $
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