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(1/5)(2x-1)=3
We move all terms to the left:
(1/5)(2x-1)-(3)=0
Domain of the equation: 5)(2x-1)!=0We add all the numbers together, and all the variables
x∈R
(+1/5)(2x-1)-3=0
We multiply parentheses ..
(+2x^2+1/5*-1)-3=0
We multiply all the terms by the denominator
(+2x^2+1-3*5*-1)=0
We get rid of parentheses
2x^2+1-1-3*5*=0
We add all the numbers together, and all the variables
2x^2=0
a = 2; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·2·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$x=\frac{-b}{2a}=\frac{0}{4}=0$
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