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n^2+18n+9=0
a = 1; b = 18; c = +9;
Δ = b2-4ac
Δ = 182-4·1·9
Δ = 288
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-12\sqrt{2}}{2*1}=\frac{-18-12\sqrt{2}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12\sqrt{2}}{2*1}=\frac{-18+12\sqrt{2}}{2} $
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