Question 401134
Let r km/h is tractor's rate from one town to another. For a distance of 180 kilometers it takes time {{{180/r}}} hours
He drives 15 kilometers per hour faster on the return trip, so it rate is (r+15) km/h. For a distance of 180 kilometers it takes {{{180/(r+15)}}} hours
 A farmer cuts one hour off the time on the return road, so
{{{180/r=180/(r+15)+1}}}
{{{(180(r+15))/(r(r+15))=(180r)/(r(r+15))+(r(r+15))/(r(r+15))}}}
{{{180(r+15)=180r+r(r+15)}}}
{{{180r+2700=180r+r^2+15r}}}
{{{r^2+15r-2700=0}}} 
a quadratic equation , the roots are given by the quadratic formula {{{r= (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{r = (-15 +- sqrt( 15^2-4*1*(-2700 )))/(2*1) }}} 
{{{r = (-15 + 105)/2 }}} or {{{r = (-15 - 105)/2 }}} 
{{{r = 45}}} or {{{r = -60 }}}<0 extraneous root of the equation, does not satisfy the problem

{{{r = 45}}}km/h is tractor's rate from one town to another, back trip rate is {{{(r+15)=45+15=60}}}km/h