Question 167947
Let {{{r}}}= the speed of the old planes Dallas-Seattle in km/hr
Let {{{t}}}= the old plane's time Dallas-Seattle in hrs
Given is
{{{d = 2800}}}km
{{{2800 = r*t}}}
{{{t = 2800/r}}}
Also given is
{{{2800 = (r + 100)*(t - .5)}}}
{{{2800 = (r + 100)*(2800/r - .5)}}}
{{{2800 = 2800 + 280000/r - .5r - 50}}}
Subtract {{{2800}}} from both sides 
{{{0 = 280000/r - .5r - 50}}}
Multiply both sides by {{{r}}}
{{{280000 - .5r^2 - 50r = 0}}}
Divide both sides by {{{.5}}}
{{{-r^2 - 100r + 560000 = 0}}}
Using the quadratic formula
{{{r = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{r = (-(-100) +- sqrt( (-100)^2-4*(-1)*560000 ))/(2*(-1)) }}}
{{{r = (100 +- sqrt( 10000 + 2240000 ))/-2) }}}
{{{r = (100 +- sqrt( 2250000 ))/-2) }}}
{{{r = (100 + 1500) / -2}}} reject this since it is negative
{{{r = (100 - 1500) / -2}}}
{{{r = -1400 / -2}}}
{{{r = 700}}}
The speed of the new planes is {{{r + 100 = 800}}} km/hr
check answer:
{{{t = 2800/r}}}
{{{t = 2800/700}}}
{{{t = 4}}}hrs
{{{2800 = (r + 100)*(t - .5)}}}
{{{2800 = (700 + 100)*(4 - .5)}}}
{{{2800 = 800*3.5}}}
{{{2800 = 2800}}}
OK