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12+1/3x+2/3x=x
We move all terms to the left:
12+1/3x+2/3x-(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+2/3x+12=0
We multiply all the terms by the denominator
-1x*3x+12*3x+1+2=0
We add all the numbers together, and all the variables
-1x*3x+12*3x+3=0
Wy multiply elements
-3x^2+36x+3=0
a = -3; b = 36; c = +3;
Δ = b2-4ac
Δ = 362-4·(-3)·3
Δ = 1332
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1332}=\sqrt{36*37}=\sqrt{36}*\sqrt{37}=6\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6\sqrt{37}}{2*-3}=\frac{-36-6\sqrt{37}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{37}}{2*-3}=\frac{-36+6\sqrt{37}}{-6} $
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