AS MATHS--Please explain how the mark scheme got this answer

The point E has coordinates (5,k) . Given that CE has length 5, find the two
possible values of the constant k.

So I did
(5 – 2)^2 + (k + 7)^2 = 5^2
9+k^2+14k+49=25
to get k^2+14k+15=0 to k=-15 or k=-1
but the mark scheme says k=-3 or k=-11

please explain, I really dont understand, thank you

If we are to help you we need to know the coordinates of C.

I'll guess that C is (2, -7).
So the vertical distance between C and E is k - -7, that's k + 7.
(5 - 2)² + (k + 7)² = 5²
9 + k² -+14k + 49 = 25
k² +14k + 33 = 0
(k + 11)(k + 3) = 0
Either k + 11 = 0 or k + 3 = 0
k = -11 or k = -3

From your answer, the point c is (2, -7) (the only reason for 2 and 7 to appear in your equation,

So, you know that point C is on a circle centered at E
Thus the equation of the circle is
(x - 5)^2 + (y -k)^2 = 25
Since the point C is on the circle
(2- 5)^2 + (-7 - k)^2 = 25
so
9 + k^2 + 14k +49 = 25
==> k^2 + 14k + 33 = 0 (where you went wrong)
==>(k +3)(k+11) = 0
==> k = -3 or k = -11

Where ddid you get '2' and '7' from?????

Assuming the point 'C' has coordinates (2,-7)
Then
9 + k^2 + 14k + 49 = 25
k^2 + 14k + 33 = 0
Factorise
(k + 3)(k+ 11) = 0
k = -3
& k = -11

NB
It would appear that
1. You have used your displacements inside the brackets incorrectly,
2. Your addition after multiplying out is incorrect.

( + 49 = 58
58 - 25 = 33
NOT '15'.

You failed to give us the coordinates of point C. Is it (2, -7)?

After 9+k^2+14k+49=25, the next equation should be
k^2 + 14k + 33 = 0 <== you had +15 here.

+9 +49 -25 = 33, not 15