How do we get k<-1, k>2 from (k+1)(k-2)>0

Note the condition using the discriminant for two distinct roots is:

$\displaystyle b^{2} > 4ac.$

(Source: https://www.mathwords.com/d/discriminant_quadratic.htm ).
Then as you have pointed out, you obtain equation:

$(k+1) (k-2) > 0, \quad \ast.$

Notice this is a quadratic which is concave up / smiley face.
For which value of k is the equation * true (simple to read off from equation * as it is already factorised) ?

I have to find the range of k for which each quadratic equation has two different positive solutions.

We are given the quadratic equation Let f(x) = x^2-2kx+k+2

Using the Discrminant D/4 = b^2-ac,
How do we get k<-1, k>2 from (k+1)(k-2)>0??

I just can't seem to work that out???

That is really appreciated. 3 people couldn't explain it to me as simple as you have.
Cheers