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2x+1/3x=350
We move all terms to the left:
2x+1/3x-(350)=0
Domain of the equation: 3x!=0We multiply all the terms by the denominator
x!=0/3
x!=0
x∈R
2x*3x-350*3x+1=0
Wy multiply elements
6x^2-1050x+1=0
a = 6; b = -1050; c = +1;
Δ = b2-4ac
Δ = -10502-4·6·1
Δ = 1102476
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1102476}=\sqrt{4*275619}=\sqrt{4}*\sqrt{275619}=2\sqrt{275619}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1050)-2\sqrt{275619}}{2*6}=\frac{1050-2\sqrt{275619}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1050)+2\sqrt{275619}}{2*6}=\frac{1050+2\sqrt{275619}}{12} $
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