Unfortunately, there are no two such numbers.
If your original question was: [tex]\sf{x^{2} -9x+24=0}[/tex] and you were trying to factor it, I would suggest the quadratic formula.
There are two solutions to [tex]\sf{ax^2+bx+c=0[/tex], where [tex]\sf{a\neq 0}[/tex], and we can find them by using the quadratic formula:
[tex]\huge{\sf{x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}}}[/tex]
So using that, here is how it would get factored:
For your equation, [tex]\sf{a = 1, b = -9, c = 24}[/tex]
Plug in those values into the formula:
[tex]\sf{\frac{ -(-9) \pm \sqrt{ (-9)^2 - 4 \cdot 1 \cdot 24} }{ 2 \cdot 1 }}[/tex]
Simplifying it:
[tex]\sf{\frac{ 9 \pm \sqrt{ -15 } }{ 2 }}[/tex]
And so the solutions are:
[tex]\sf{x_1 = \frac{ 9~+~\sqrt{ -15 } }{ 2 } = \frac{9}{2}+\frac{1}{2}\sqrt{15}i}[/tex]
and
[tex]\sf{x_2 = \frac{ 9~-~\sqrt{ -15 } }{ 2 } = \frac{9}{2}-\frac{1}{2}\sqrt{15}i}[/tex]
