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48n^2-19n+1=0
a = 48; b = -19; c = +1;
Δ = b2-4ac
Δ = -192-4·48·1
Δ = 169
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{169}=13$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*48}=\frac{6}{96} =1/16 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*48}=\frac{32}{96} =1/3 $
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