Let $r$ and $s$ be the roots of $y^2 - 19y + 7$. Find $(r-2)(s-2)$.
y^2 - 19y + 7 (r - 2) (s - 2) = rs - 2s - 2r + 4 = rs -2(r + s) + 4
Using Vieta's Theorem
The sum of the roots , (r + s) = 19
The product of the roots, rs = 7
So
(r - 2) (s - 2) = rs - 2(r + s) + 4 = 7 - 2(19) + 4 = -27
