First of all, we have to find the roots of the polynomial:
[tex]r^2-2r-15=0 \iff r=\dfrac{2\pm\sqrt{4+60}}{2}=\dfrac{2\pm 8}{2}=1\pm 4[/tex]
So, the two solutions are
[tex]r_1=1+4=5,\quad r_2=1-4=-3[/tex]
And we can use them to factor the polynomial as
[tex]r^2-2r-15=(r+3)(r-5)[/tex]
So, we have written the area as the product of two dimensions, which means that the length and the width of the rectangles are [tex]r+3[/tex] and [tex]r-5[/tex].
This implies that [tex]r[/tex] has to be at least 5, otherwise you'd have negative dimensions.
