Question 959641
Let {{{ s }}} = the usual speed in mi/hr that will result 
in the bus arriving on time
Let {{{ t }}} = the time in hours that the trip takes
when the bus has speed {{{ s }}}
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Equation for a normal trip:
(1) {{{ 330 = s*t }}}
Equation for trip leaving at 5:30 PM
(2) {{{ 330 = ( s + 5 )*( t - .5 ) }}} 
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(2) {{{ 330 = s*t + 5t - .5s - 2.5 }}}
(2) {{{ 332.5 = s*t + 5t - .5s }}}
and
(1) {{{ s = 330/t }}}
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By substitution:
(2) {{{ 332.5  = 330 + 5t - .5*( 330/t ) }}}
(2) {{{ 2.5 = 5t - 165/t }}}
(2) {{{ 5t^2 - 2.5t - 165 = 0 }}}
Divide both sides by {{{ 2.5 }}}
(2) {{{ 2t^2 - t - 66 = 0 }}}
Use quadratic formula
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 2 }}}
{{{ b = -1 }}}
{{{ c = -66 }}}
{{{ t = ( -(-1) +- sqrt( (-1)^2 - 4*2*(-66) )) / (2*2) }}}
{{{ t = ( 1 +- sqrt( 1 + 528 )) / 4 }}}
{{{ t = ( 1 +- sqrt( 529 )) / 4 }}}
{{{ t = ( 1 + 23 ) / 4 }}}
{{{ t = 24/4 }}}
{{{ t = 6 }}}
and
(1) {{{ s = 330/t }}}
(1) {{{ s = 330/6 }}}
(1) {{{ s = 55 }}}
The usual speed is 55 mi/hr
check:
(2) {{{ 330 = ( s + 5 )*( t - .5 ) }}} 
(2) {{{ 330 = ( 55 + 5 )*( 6 - .5 ) }}} 
(2) {{{ 330 = 60*5.5 }}}
(2) {{{ 330 = 330 }}}
OK