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n^2+2n-165=0
a = 1; b = 2; c = -165;
Δ = b2-4ac
Δ = 22-4·1·(-165)
Δ = 664
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{664}=\sqrt{4*166}=\sqrt{4}*\sqrt{166}=2\sqrt{166}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{166}}{2*1}=\frac{-2-2\sqrt{166}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{166}}{2*1}=\frac{-2+2\sqrt{166}}{2} $
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