Question 1080617
Initial table of data before analyzing the time quantities; w is tailwind speed.
<pre>
               SPEED        TIME         DISTANCE

WITHWIND       155+w                     950
AGAINSTWD      155-w                     950
</pre>

<pre>
               SPEED        TIME            DISTANCE

WITHWIND       155+w        950/(155+w)      950
AGAINSTWD      155-w        950/(155-w)      950
</pre>

Concentrate on the part of the description, <i>Flying against a headwind, it takes 
3 hours longer to complete the return trip.</i>
This means {{{highlight(highlight_green(950/(155-w)-950/(155+w)=3))}}}.


The rest is arithmetic to solve the equation for {{{w}}}, the speed of the wind.


{{{ 950/(155-w)-950/(155+w)=3}}}


{{{ (950(155+w)-950(155-w))/(155-w)(155+w)=3}}}


{{{ (950*155+950w-950*155+950w)/(155^2-w^2)=3}}}


{{{950w+950w=3(155^2-w^2)}}}  
 

{{{1900w=3*24025-3w^2}}} 


{{{1900w=72075-3w^2}}} 


{{{3w^2+1900w-72075=0}}}......using {{{w = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} you will get: 


 {{{w = -950/3 - (5*sqrt(44749))/3}}}

{{{w =  - 950/3+(5*sqrt(44749))/3}}}