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2/5x+2(5/3x+2)=4
We move all terms to the left:
2/5x+2(5/3x+2)-(4)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 3x+2)!=0We multiply parentheses
x∈R
2/5x+10x+4-4=0
We multiply all the terms by the denominator
10x*5x+4*5x-4*5x+2=0
Wy multiply elements
50x^2+20x-20x+2=0
We add all the numbers together, and all the variables
50x^2+2=0
a = 50; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·50·2
Δ = -400
Delta is less than zero, so there is no solution for the equation
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