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

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

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

We move all terms to the left:

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

We use the square of the difference formula

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+t^2-3t+2t-6)+4), so:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

t^2-1t-2

Back to the equation:

-(t^2-1t-2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

t+1=0

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

t=-1