(t+2)(t-2)=(t+5)(t-6)+8

Solution for (t+2)(t-2)=(t+5)(t-6)+8 equation:

(t+2)(t-2)=(t+5)(t-6)+8

We move all terms to the left:

(t+2)(t-2)-((t+5)(t-6)+8)=0

We use the square of the difference formula

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

We multiply parentheses ..

t^2-((+t^2-6t+5t-30)+8)-4=0


We calculate terms in parentheses: -((+t^2-6t+5t-30)+8), so:

(+t^2-6t+5t-30)+8

We get rid of parentheses

t^2-6t+5t-30+8

We add all the numbers together, and all the variables

t^2-1t-22

Back to the equation:

-(t^2-1t-22)


We get rid of parentheses

t^2-t^2+1t+22-4=0

We add all the numbers together, and all the variables

t+18=0

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

t=-18