Here is a start:
(r-2)(s-2).= rs -2r-2s+4
= rs -2 (r+s) +4 Are you familiar with Vieta's formulas for quadratics???
Vieta's for quadratics : r + s = -b/a =19
and rs = c/a = 9
rs -2 (r+s) +4 =
9 -2(19)+4 = -25
For this, we don't ACTUALLY have to solve for r and s.
Let us first expand (r-2)(s-2):
if r and s are the roots of y^2 - 19y + 9, then:
$y^2 - 19y + 9 = (y-r)(y-s) = y^2 - (r + s)y + rs$
so r+s=19 and rs=9
going back to the original equation, we have