n2=18n+40

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Solution for n2=18n+40 equation:



n2=18n+40
We move all terms to the left:
n2-(18n+40)=0
We add all the numbers together, and all the variables
n^2-(18n+40)=0
We get rid of parentheses
n^2-18n-40=0
a = 1; b = -18; c = -40;
Δ = b2-4ac
Δ = -182-4·1·(-40)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*1}=\frac{-4}{2} =-2 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*1}=\frac{40}{2} =20 $

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