Hi,
One place to start is to use the basic quadratic formula:
x=(-b+/-(b^2-4ac)^0.5)/2a.
With a=1, b=-10, and c=k, and letting x=5 (as one of the roots) gives you k=25.
Substituting this back into the question gives you x^2-10x+25=0. This is easily factorised to give x=5, twice.
So the answer to the question is that the other solution is also 5.
you can use discriminant X(squared)-11x+k=aX(squared)+bx+c
a=1, b=-11, c=k, and let's say the discriminant D
D=b(squared)-4ac
=121-4*1*k
=121-4k
solutions are s1 and s2
S1=(-b+(square root)D)/2a
=(-11+(square root)121-4k)/2
and S1=5...so
(-11+(square root)121-4k)/2=5
-11+(square root)121-4k=10
(square root)121-4k=21
now you put everywhere by squared and you get
121-4k=441
-4k=320
k=-80
so now let's look for S2
D=(square root)121-4(-80)
D=(square root)121+320
D=(square root)441
D=21
S2=(-b-D)/2a
S2=(-11-21)/2
S2=-16
the second root is -16