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(1/5)x+3=2
We move all terms to the left:
(1/5)x+3-(2)=0
Domain of the equation: 5)x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
(+1/5)x+3-2=0
We add all the numbers together, and all the variables
(+1/5)x+1=0
We multiply parentheses
x^2+1=0
a = 1; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·1·1
Δ = -4
Delta is less than zero, so there is no solution for the equation
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