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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((2t-3)(t+2)), so:

(2t-3)(t+2)

We multiply parentheses ..

(+2t^2+4t-3t-6)

We get rid of parentheses

2t^2+4t-3t-6

We add all the numbers together, and all the variables

2t^2+t-6

Back to the equation:

-(2t^2+t-6)


We get rid of parentheses

2t^2-2t^2+12t+t-t+6+6=0

We add all the numbers together, and all the variables

12t+12=0

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

12t=-12

t=-12/12

t=-1