Question 806302
Given:
{{{ A = 33 }}} m2
{{{ L = 2W -5 }}} m
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The formula for area is
{{{ A = L*W }}}
By substitution:
{{{ A = ( 2W - 5 )*W }}}
{{{ 33 = 2W^2 - 5W }}}
{{{ 2W^2 - 5W - 33 = 0 }}}
Using the quadratic formula:
{{{ W  = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 2 }}}
{{{ b = -5 }}}
{{{ c = -33 }}}
{{{ W  = (-(-5) +- sqrt( (-5)^2 - 4*2*(-33) )) / (2*2) }}}
{{{ W  = ( 5 +- sqrt( 25 + 264 )) / 4 }}}
{{{ W  = ( 5 +- sqrt( 289 )) / 4 }}} ( I can't use the negative square root )
{{{ W = ( 5 + 17 ) / 4 }}} 
{{{ W = 22/4 }}}
{{{ W = 11/2 }}}
and, since
{{{ L = 2W -5 }}} 
{{{ L = 2*(11/2) -5 }}} 
{{{ L = 11 - 5 }}}
{{{ L = 6 }}}
The rectangle is 6 m x 5.5 m
check:
{{{ A = L*W }}}
{{{ A = 6*5.5 }}}
{{{ 33 = 33 }}}
OK