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n^2-18n+66=0
a = 1; b = -18; c = +66;
Δ = b2-4ac
Δ = -182-4·1·66
Δ = 60
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{15}}{2*1}=\frac{18-2\sqrt{15}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{15}}{2*1}=\frac{18+2\sqrt{15}}{2} $
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