Question 292167
Let n = the required number, then:
{{{n^2 = 3n+28}}} Rearrange into the standard form for a quadratic equation.
{{{n^2-3n-28 = 0}}} Factor this quadratic equation.
{{{(n+4)(n-7) = 0}}} Apply the "zero product" rule.
{{{n+4 = 0}}} or {{{n-7 = 0}}} so that...
{{{highlight(n = -4)}}} or {{{highlight(n = 7)}}}
So there are two answers to this problem.  Let's check them both:
{{{n^2 = 3n+28}}} Substitute n = -4.
{{{(-4)^2 = 3(-4)+28}}}
{{{16 = -12+28}}}
{{{16 = 16}}} OK for n = -3.
{{{n^2 = 3n+28}}} Substitute n = 7.
{{{7^2 = 3(7)+28}}}
{{{49 = 21+28}}}
{{{49 = 49}}} OK for n = 7.