`f(t) = t^2+5t-7`

a)

`f(t) = -1`

`t^2+5t-7 = -1`

`t^2+5t-6 = 0`

`t^2+6t-t-6 = 0`

`t(t+6)-1(t+6) = 0`

`(t+6)(t-1) = 0`

`t = 1` or `t = -6`

*So the answers are t = 1 and t = -6.*

b)

to find f(k+3) we need to replace t with (k+3).

`f(t)=t^2+5t-7`

`f(k+3) = (k+3)^2+5(k+3)-7`

`f(k+3) = k^2+11k+17`

*So the answer is `f(k+3) = k^2+11k+17` .*

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