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n^2-14n+43=0
a = 1; b = -14; c = +43;
Δ = b2-4ac
Δ = -142-4·1·43
Δ = 24
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{6}}{2*1}=\frac{14-2\sqrt{6}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{6}}{2*1}=\frac{14+2\sqrt{6}}{2} $
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