1/3x+0.2x=16

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Solution for 1/3x+0.2x=16 equation:

1/3x+0.2x=16
We move all terms to the left:
1/3x+0.2x-(16)=0

Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R

We add all the numbers together, and all the variables
0.2x+1/3x-16=0
We multiply all the terms by the denominator

(0.2x)*3x-16*3x+1=0
We add all the numbers together, and all the variables
(+0.2x)*3x-16*3x+1=0
We multiply parentheses
0x^2-16*3x+1=0
Wy multiply elements
0x^2-48x+1=0
We add all the numbers together, and all the variables
x^2-48x+1=0
a = 1; b = -48; c = +1;Δ = b2-4acΔ = -482-4·1·1Δ = 2300The delta value is higher than zero, so the equation has two solutionsWe use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$The end solution:
$\sqrt{\Delta}=\sqrt{2300}=\sqrt{100*23}=\sqrt{100}*\sqrt{23}=10\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-10\sqrt{23}}{2*1}=\frac{48-10\sqrt{23}}{2}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+10\sqrt{23}}{2*1}=\frac{48+10\sqrt{23}}{2}$

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