2x+4/3x+20x=180

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Solution for 2x+4/3x+20x=180 equation:



2x+4/3x+20x=180
We move all terms to the left:
2x+4/3x+20x-(180)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
We add all the numbers together, and all the variables
22x+4/3x-180=0
We multiply all the terms by the denominator
22x*3x-180*3x+4=0
Wy multiply elements
66x^2-540x+4=0
a = 66; b = -540; c = +4;
Δ = b2-4ac
Δ = -5402-4·66·4
Δ = 290544
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:
$\sqrt{\Delta}=\sqrt{290544}=\sqrt{16*18159}=\sqrt{16}*\sqrt{18159}=4\sqrt{18159}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-540)-4\sqrt{18159}}{2*66}=\frac{540-4\sqrt{18159}}{132} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-540)+4\sqrt{18159}}{2*66}=\frac{540+4\sqrt{18159}}{132} $

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