SOLUTION: The roots of x²+kx+6=0 are each 5 less than the roots of x²−kx+6=0. Find k. Thanks so much for helping!

Let the roots of the second equation x²-kx+6 = 0 be r and s. Then since
the leading coefficient is 1, the sum of the roots equals the coefficient
of x with the sign changed, so,

r+s = k

Then the roots of the first equation x²+kx+6 = 0 are r-5 and s-5. Then since
the leading coefficient is 1, the sum of the roots equals the coefficient
of x with the sign changed, so,

r-5 + s-5 = -k, which simplifies to
r+s-10 = -k

So we have the system:
r+s = k
r+s-10 = -k

Substitute k for r+s in the second:

k-10 = -k
  2k = 10
   k = 5

Checking:

Solve:

   x²+5x+6 = 0           x²-5x+6 = 0
(x+3)(x+2) = 0        (x-3)(x-2) = 0   
 x+3=0; x+2=0          x-3=0; x-2=0 
  x=-3;   x=-2           x=3;   x=2

the root -3 of the first equation is 5 less than the root 2 of the
second equation, since -3 = 2-5.  Also:
the root -2 of the first equation is 5 less than the root 3 of the
second equation, since -2 = 3-5. 
 
Edwin