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(14+1/3x+x)=65
We move all terms to the left:
(14+1/3x+x)-(65)=0
Domain of the equation: 3x+x)!=0We add all the numbers together, and all the variables
x∈R
(x+1/3x+14)-65=0
We get rid of parentheses
x+1/3x+14-65=0
We multiply all the terms by the denominator
x*3x+14*3x-65*3x+1=0
Wy multiply elements
3x^2+42x-195x+1=0
We add all the numbers together, and all the variables
3x^2-153x+1=0
a = 3; b = -153; c = +1;
Δ = b2-4ac
Δ = -1532-4·3·1
Δ = 23397
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-153)-\sqrt{23397}}{2*3}=\frac{153-\sqrt{23397}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-153)+\sqrt{23397}}{2*3}=\frac{153+\sqrt{23397}}{6} $
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