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n^2-2n-47=0
a = 1; b = -2; c = -47;
Δ = b2-4ac
Δ = -22-4·1·(-47)
Δ = 192
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8\sqrt{3}}{2*1}=\frac{2-8\sqrt{3}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8\sqrt{3}}{2*1}=\frac{2+8\sqrt{3}}{2} $
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