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16n^2-26n+3=0
a = 16; b = -26; c = +3;
Δ = b2-4ac
Δ = -262-4·16·3
Δ = 484
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{484}=22$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-22}{2*16}=\frac{4}{32} =1/8 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+22}{2*16}=\frac{48}{32} =1+1/2 $
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