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1/3x+12=x
We move all terms to the left:
1/3x+12-(x)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
-1x+1/3x+12=0
We multiply all the terms by the denominator
-1x*3x+12*3x+1=0
Wy multiply elements
-3x^2+36x+1=0
a = -3; b = 36; c = +1;
Δ = b2-4ac
Δ = 362-4·(-3)·1
Δ = 1308
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:x_{1}=\frac{-b-\sqrt{\Delta}}{2a}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}
The end solution:
\sqrt{\Delta}=\sqrt{1308}=\sqrt{4*327}=\sqrt{4}*\sqrt{327}=2\sqrt{327}x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{327}}{2*-3}=\frac{-36-2\sqrt{327}}{-6}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{327}}{2*-3}=\frac{-36+2\sqrt{327}}{-6}
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