(3k+4)(9k+5)=(3k+4)(9k+5)

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Solution for (3k+4)(9k+5)=(3k+4)(9k+5) equation: 



(3k+4)(9k+5)=(3k+4)(9k+5)

We move all terms to the left:

(3k+4)(9k+5)-((3k+4)(9k+5))=0

We multiply parentheses ..

(+27k^2+15k+36k+20)-((+27k^2+15k+36k+20))=0



We calculate terms in parentheses: -((+27k^2+15k+36k+20)), so:

(+27k^2+15k+36k+20)

We get rid of parentheses

27k^2+15k+36k+20

We add all the numbers together, and all the variables

27k^2+51k+20

Back to the equation:

-(27k^2+51k+20)



We get rid of parentheses

27k^2-27k^2+15k+36k-51k+20-20=0

We add all the numbers together, and all the variables

=0

k=0/1

k=0

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