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n^2+18=11n
We move all terms to the left:
n^2+18-(11n)=0
a = 1; b = -11; c = +18;
Δ = b2-4ac
Δ = -112-4·1·18
Δ = 49
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-7}{2*1}=\frac{4}{2} =2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+7}{2*1}=\frac{18}{2} =9 $
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