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(5/6)-(2/3)n=(3/2)
We move all terms to the left:
(5/6)-(2/3)n-((3/2))=0
Domain of the equation: 3)n!=0We add all the numbers together, and all the variables
n!=0/1
n!=0
n∈R
-(+2/3)n+(+5/6)-((+3/2))=0
We multiply parentheses
-2n^2+(+5/6)-((+3/2))=0
We get rid of parentheses
-2n^2+5/6-((+3/2))=0
We calculate fractions
-2n^2+()/()+()/()=0
We add all the numbers together, and all the variables
-2n^2+2=0
a = -2; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-2)·2
Δ = 16
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{16}=4$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-2}=\frac{-4}{-4} =1 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-2}=\frac{4}{-4} =-1 $
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