The equations of pairs of opposite sides of a parallelogram are $x^2-5x+6=0$  and $y^2-6y+5=0$ . The equations of its diagonals are

The two pairs are $x^2-5x+6=0$  and $y^2-6y+5=0$ . Hence, the sides of the parallelogram are , $y=1$ , $x=3,y=5,$ , $x=2$ . The vertices, taken in order are $A(2,1),B(3,1),C(3,5)$  and $D(2,5)$ . Hence

The two diagonals are $\left(ACandBD\right)$

         $y-1=\frac{5-1}{3-2}(x-2)$

         $y-1=4(x-2)$

And   $y-1=\frac{5-1}{2-3}(x-3)$

$y-1=\frac{4}{-1}\left(x-3\right)$

i.e. $4x-y-7=0$  and $4x+y-13=0$ .