Find a lower bound of $k$ when $m!=100x^2+20x$ and $k*100x^2+20x=(m+5)!$

Question

$x$ & $m$ & $k$ are all real non-negative numbers. (Not necessarily intergers)

I have the equation: $m!=100x^2+20x$

$100x^2+20x\neq (m+5)!$

$k*100x^2+20x=(m+5)!$ Can some one write a lower bound of an approximation for $k$ using $x$ and/or $m$.

Answer 1

Since $$(m+5)!=m! × (m+1)(m+2)(m+3)(m+4)(m+5)$$ $$k=(m+1)(m+2)(m+3)(m+4)(m+5)$$ or by estimation $ k>m^5$