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n^2-14n-50=0
a = 1; b = -14; c = -50;
Δ = b2-4ac
Δ = -142-4·1·(-50)
Δ = 396
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6\sqrt{11}}{2*1}=\frac{14-6\sqrt{11}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6\sqrt{11}}{2*1}=\frac{14+6\sqrt{11}}{2} $
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