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(12x^2-4x+1)/(3x+2)=x
We move all terms to the left:
(12x^2-4x+1)/(3x+2)-(x)=0
Domain of the equation: (3x+2)!=0We add all the numbers together, and all the variables
We move all terms containing x to the left, all other terms to the right
3x!=-2
x!=-2/3
x!=-2/3
x∈R
-1x+(12x^2-4x+1)/(3x+2)=0
We multiply all the terms by the denominator
-1x*(3x+2)+(12x^2-4x+1)=0
We multiply parentheses
-3x^2-2x+(12x^2-4x+1)=0
We get rid of parentheses
-3x^2+12x^2-2x-4x+1=0
We add all the numbers together, and all the variables
9x^2-6x+1=0
a = 9; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·9·1
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$x=\frac{-b}{2a}=\frac{6}{18}=1/3$
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