x² - (2k-5)x + (k²-5k+56/9) = 0

Use these facts of quadratics with leading coefficient 1:
For the quadratic equation x²+Ax+B=0
The sum of the roots  is -A
The product of the roots is B

Let r = the root that is NOT twice the other

Then the root which IS twice the other is 2r

Sum of roots = r+2r = 3r

Therefore 

         3r = 2k-5

and       r = 

That was the first thing you were to show.

Product of roots = (r)(2r) = 2r²

Therefore 

        2r² = k²-5k+

So we have the system of equations:

1st:     3r = 2k-5
2nd:    2r² = k²-5k+

Clear the 2nd equation of fractions:

       18r² = 9k²-45k+56

Square both sides of the 1st

        9r² = 4k²-20k+25

Multiply through by 2

       18r² = 8k²-40k+50

Set the expressions for 18x² equal:

 9k²-45k+56 = 8k²-40k+50

k² - 5k + 6 = 0
 (k-3)(k-2) = 0

k-3=0;  k-2=0
  k=3;    k=2

So those are the 2 values of k, which is what was asked for.

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We weren't asked for the roots, but we could find the roots of the given
 equation very easily:

for k = 3, substitute in

3r = 2k-5
3r = 2(3)-5
3r = 6-5
3r = 1
 r = 

So for k = 3, one root is x =  and the other is x =  

for k = 2, substitute in

3r = 2k-5
3r = 2(2)-5
3r = 4-5
3r = -1
 r = 

So for k = 2, one root is x =  and the other is x = 

Edwin