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n-5/3n+3=0
Domain of the equation: 3n!=0We multiply all the terms by the denominator
n!=0/3
n!=0
n∈R
n*3n+3*3n-5=0
Wy multiply elements
3n^2+9n-5=0
a = 3; b = 9; c = -5;
Δ = b2-4ac
Δ = 92-4·3·(-5)
Δ = 141
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{-(9)-\sqrt{141}}{2*3}=\frac{-9-\sqrt{141}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{141}}{2*3}=\frac{-9+\sqrt{141}}{6} $
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