(3t-2)(4t+1)=(6t-5)(2t+4)

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Solution for (3t-2)(4t+1)=(6t-5)(2t+4) equation: 



(3t-2)(4t+1)=(6t-5)(2t+4)

We move all terms to the left:

(3t-2)(4t+1)-((6t-5)(2t+4))=0

We multiply parentheses ..

(+12t^2+3t-8t-2)-((6t-5)(2t+4))=0



We calculate terms in parentheses: -((6t-5)(2t+4)), so:

(6t-5)(2t+4)

We multiply parentheses ..

(+12t^2+24t-10t-20)

We get rid of parentheses

12t^2+24t-10t-20

We add all the numbers together, and all the variables

12t^2+14t-20

Back to the equation:

-(12t^2+14t-20)



We get rid of parentheses

12t^2-12t^2+3t-8t-14t-2+20=0

We add all the numbers together, and all the variables

-19t+18=0

We move all terms containing t to the left, all other terms to the right

-19t=-18

t=-18/-19

t=18/19

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