2(1/3k-4)^2+5/3k-4+2=0

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Solution for 2(1/3k-4)^2+5/3k-4+2=0 equation: 



2(1/3k-4)^2+5/3k-4+2=0



Domain of the equation: 3k-4)^2!=0

k∈R





Domain of the equation: 3k!=0

k!=0/3

k!=0

k∈R



We add all the numbers together, and all the variables

2(1/3k-4)^2+5/3k-2=0

We calculate fractions


(2(1*3k)/(3k-4)^2*3k)+15k/(3k-4)^2*3k)-2=0



We calculate terms in parentheses: +(2(1*3k)/(3k-4)^2*3k), so:

2(1*3k)/(3k-4)^2*3k

We add all the numbers together, and all the variables

2(+1*3k)/(3k-4)^2*3k

We multiply all the terms by the denominator


2(+1*3k)

We multiply parentheses

6k

Back to the equation:

+(6k)



We multiply all the terms by the denominator


6k*(3k-4)^2*3k)-2+15k=0

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