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n^2-26n+168=0
a = 1; b = -26; c = +168;
Δ = b2-4ac
Δ = -262-4·1·168
Δ = 4
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{4}=2$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2}{2*1}=\frac{24}{2} =12 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2}{2*1}=\frac{28}{2} =14 $
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