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n^2-18n-44=0
a = 1; b = -18; c = -44;
Δ = b2-4ac
Δ = -182-4·1·(-44)
Δ = 500
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-10\sqrt{5}}{2*1}=\frac{18-10\sqrt{5}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+10\sqrt{5}}{2*1}=\frac{18+10\sqrt{5}}{2} $
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