The quadrilateral formed by the lines whose equations are given by $x^2-5x+6=0$ and $y^2-14y+40=0$ is

$\text{ The given lines are }{x}^2-5x+6=0\text{ and }{y}^2-14y+40=0$

$\text{ It implies that the equations of sides are quadrilateral are }$

$x-1=0,x-5=0,y-10=0,y-4=0$

The sides of quadrilateral are vertical and horizontal lines, so angles are right angles

Distances between corresponding opposite sides are $5-1=4,10-4=6$

Since the distances between two parallel sides are not equal, and the angles are right angles, the figure formed by the lines is rectangle.