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(t+3)(t-3)=(t+4)(t-5)+6
We move all terms to the left:
(t+3)(t-3)-((t+4)(t-5)+6)=0
We use the square of the difference formula
t^2-((t+4)(t-5)+6)-9=0
We multiply parentheses ..
t^2-((+t^2-5t+4t-20)+6)-9=0
We calculate terms in parentheses: -((+t^2-5t+4t-20)+6), so:We get rid of parentheses
(+t^2-5t+4t-20)+6
We get rid of parentheses
t^2-5t+4t-20+6
We add all the numbers together, and all the variables
t^2-1t-14
Back to the equation:
-(t^2-1t-14)
t^2-t^2+1t+14-9=0
We add all the numbers together, and all the variables
t+5=0
We move all terms containing t to the left, all other terms to the right
t=-5
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