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(1/5)x+3=2x-24
We move all terms to the left:
(1/5)x+3-(2x-24)=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-(2x-24)+3=0
We multiply parentheses
x^2-(2x-24)+3=0
We get rid of parentheses
x^2-2x+24+3=0
We add all the numbers together, and all the variables
x^2-2x+27=0
a = 1; b = -2; c = +27;
Δ = b2-4ac
Δ = -22-4·1·27
Δ = -104
Delta is less than zero, so there is no solution for the equation
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