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n(n-2)=10
We move all terms to the left:
n(n-2)-(10)=0
We multiply parentheses
n^2-2n-10=0
a = 1; b = -2; c = -10;
Δ = b2-4ac
Δ = -22-4·1·(-10)
Δ = 44
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{11}}{2*1}=\frac{2-2\sqrt{11}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{11}}{2*1}=\frac{2+2\sqrt{11}}{2} $
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