-1/3z+7z=20/3z

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Solution for -1/3z+7z=20/3z equation:



-1/3z+7z=20/3z
We move all terms to the left:
-1/3z+7z-(20/3z)=0
Domain of the equation: 3z!=0
z!=0/3
z!=0
z∈R
Domain of the equation: 3z)!=0
z!=0/1
z!=0
z∈R
We add all the numbers together, and all the variables
-1/3z+7z-(+20/3z)=0
We add all the numbers together, and all the variables
7z-1/3z-(+20/3z)=0
We get rid of parentheses
7z-1/3z-20/3z=0
We multiply all the terms by the denominator
7z*3z-1-20=0
We add all the numbers together, and all the variables
7z*3z-21=0
Wy multiply elements
21z^2-21=0
a = 21; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·21·(-21)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42}{2*21}=\frac{-42}{42} =-1 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42}{2*21}=\frac{42}{42} =1 $

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