Question 50015
Let the required number(s) be n.

From the problem description, you can write:
{{{2n = n^2-8}}} Subtract 2n from both sides.
{{{n^2-2n-8 = 0}}} Solve by factoring.
{{{(n+2)(n-4) = 0}}} Apply the zero product principle.
{{{n+2 = 0}}} and/or {{{n-4 = 0}}}
If {{{n+2 = 0}}} then {{{n = -2}}}
If {{{n-4 = 0}}} then {{{n = 4}}}

There are two solutions to this problem:

n = -2 and
n = 4

Check:
n = -2
{{{2(-2) = (-2)^2-8}}}
{{{-4 = 4-8}}}
{{{-4 = -4}}}

n = 4

{{{2(4) = (4)^2-8}}}
{{{8 = 16-8}}}
{{{8 = 8}}}

Both solutions work.