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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2t^2+t-((2t-1)(t-2))=0

We multiply parentheses ..

2t^2-((+2t^2-4t-1t+2))+t=0


We calculate terms in parentheses: -((+2t^2-4t-1t+2)), so:

(+2t^2-4t-1t+2)

We get rid of parentheses

2t^2-4t-1t+2

We add all the numbers together, and all the variables

2t^2-5t+2

Back to the equation:

-(2t^2-5t+2)


We add all the numbers together, and all the variables

2t^2+t-(2t^2-5t+2)=0

We get rid of parentheses

2t^2-2t^2+t+5t-2=0

We add all the numbers together, and all the variables

6t-2=0

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

6t=2

t=2/6

t=1/3