2((1)/(3k-12))^(2)+(5)/((3k-12))+2=0

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Solution for 2((1)/(3k-12))^(2)+(5)/((3k-12))+2=0 equation: 



2((1)/(3k-12))^(2)+(5)/((3k-12))+2=0



Domain of the equation: (3k-12))^2!=0

k∈R





Domain of the equation: ((3k-12))!=0

k∈R



We calculate fractions


(2(1*((3k-12)))/((3k-12))^2*((3k-12)))+(5*(3k-12))^2)/((3k-12))^2*((3k-12)))+2=0



We calculate terms in parentheses: +(2(1*((3k-12)))/((3k-12))^2*((3k-12))), so:

2(1*((3k-12)))/((3k-12))^2*((3k-12))

We multiply all the terms by the denominator


2(1*((3k-12)))

Back to the equation:

+(2(1*((3k-12))))



We multiply all the terms by the denominator


((2(1*((3k-12)))))*((3k-12))^2*((3k-12)))+2+(5*(3k-12))^2)=0



We calculate terms in parentheses: +((2(1*((3k-12)))))*((3k-12))^2*((3k-12)))+2+(5*(3k-12))^2), so:

(2(1*((3k-12)))))*((3k-12))^2*((3k-12)))+2+(5*(3k-12))^2

We add all the numbers together, and all the variables

(2(1*((3k-12)))))*((3k-12))^2*((3k-12)))+2+(5*(3k

Back to the equation:

+((2(1*((3k-12)))))*((3k-12))^2*((3k-12)))+2+(5*(3k)

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