(2t-1)(2t+1)-(4t-3)=4t*(t-2)

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

(2t-1)(2t+1)-(4t-3)=4t*(t-2)

We move all terms to the left:

(2t-1)(2t+1)-(4t-3)-(4t*(t-2))=0

We use the square of the difference formula

4t^2-(4t-3)-(4t*(t-2))-1=0

We get rid of parentheses

4t^2-4t-(4t*(t-2))+3-1=0


We calculate terms in parentheses: -(4t*(t-2)), so:

4t*(t-2)

We multiply parentheses

4t^2-8t

Back to the equation:

-(4t^2-8t)


We add all the numbers together, and all the variables

4t^2-4t-(4t^2-8t)+2=0

We get rid of parentheses

4t^2-4t^2-4t+8t+2=0

We add all the numbers together, and all the variables

4t+2=0

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

4t=-2

t=-2/4

t=-1/2