n2+n=110

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Solution for n2+n=110 equation:



n2+n=110
We move all terms to the left:
n2+n-(110)=0
We add all the numbers together, and all the variables
n^2+n-110=0
a = 1; b = 1; c = -110;
Δ = b2-4ac
Δ = 12-4·1·(-110)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-21}{2*1}=\frac{-22}{2} =-11 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+21}{2*1}=\frac{20}{2} =10 $

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