Question 903851
Let {{{ t }}} = the bus's normal time  
in hrs to make the trip
{{{ 270/t }}} = the bus's normal speed
---------------------------------
{{{ 270 = ( 270/t + 5 )*( t - .5 ) }}}
{{{ 270 = 270 + 5t - 135/t - 2.5 }}}
{{{ 0 = 5t - 135/t - 2.5 }}}
{{{ 5t - 2.5 - 135/t = 0 }}}
{{{ 5t^2 - 2.5t - 135 = 0 }}}
{{{ 50t^2 - 25t - 1350 = 0 }}}
{{{ 2t^2 - t - 54 = 0 }}}
----------------------
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 2 }}}
{{{ b = -1 }}}
{{{ c = -54 }}}
{{{ t = ( -(-1) +- sqrt( (-1)^2 - 4*2*(-54) )) / (2*2) }}}
{{{ t = ( 1 +- sqrt( 1 + 432 )) / 4 }}}
{{{ t = ( 1 +- sqrt( 433 )) / 4 }}}
{{{ t = ( 1 + 20.809 ) / 4 }}}
{{{ t = 21.809 / 4 }}}
{{{ t = 5.452 }}}
{{{ 270/t }}} = the bus's normal speed
{{{ 270/5.452 = 49.523 }}}
The bus's normal speed was 49.523 mi/hr
------------
check:
{{{ 270 = ( 270/t + 5 )*( t - .5 ) }}}
{{{ 270 = ( 270/5.452 + 5 )*( 5.452 - .5 ) }}}
{{{ 270 = ( 49.523 + 5 )*4.952 }}}
{{{ 270 = 54.523*4.952 }}}
{{{ 270 = 269.9998 }}}
close enough
Hope I got it