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(1/3)n=10-n
We move all terms to the left:
(1/3)n-(10-n)=0
Domain of the equation: 3)n!=0We add all the numbers together, and all the variables
n!=0/1
n!=0
n∈R
(+1/3)n-(-1n+10)=0
We multiply parentheses
n^2-(-1n+10)=0
We get rid of parentheses
n^2+1n-10=0
We add all the numbers together, and all the variables
n^2+n-10=0
a = 1; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·1·(-10)
Δ = 41
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{41}}{2*1}=\frac{-1-\sqrt{41}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{41}}{2*1}=\frac{-1+\sqrt{41}}{2} $
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