Question 401529
Let x is the width of a rectangle, 
The length of a rectangle is  10m less than twice the width, so the lenght is {{{(2x-10)}}} m
The area of rectangle is {{{x*(2x-10)}}}, from text the area is 72m^2
equation
{{{x*(2x-10)=72}}}
{{{2x^2-10x=72}}}
{{{2x^2-10x-72=0}}} divide both sides by 2
{{{x^2-5x-36=0}}} quadratic equation
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-(-5) +- sqrt( (-5)^2-4*1*(-36)))/(2*1) }}} 
{{{x = (5 + sqrt( 169))/2 }}} or {{{x = (5- sqrt( 169))/2 }}} 
{{{x = (5 + 13)/2 }}} or {{{x = (5- 13)/2 }}} 
{{{x = 18/2 }}} or {{{x = - 8/2 }}} 
{{{x =9 }}} or {{{x = - 4 }}}<0,extraneous root of the equation does not satisfy the conditions of the problem
So width of a rectangle is 9 m, the length of a rectangle is {{{(2x-10)=(2*9-10)=8}}}m
Answer 9 m and 8 m