The roots of the equation $x(x+2)(x+3)(x+5)=72$  is

$x(x+2)(x+3)(x+5)=72$

$x(x+5)(x+2)(x+3)=72$

$(x^2+5x)(x^2+5x+6)=72$

Let $\text{Let\hspace{0.17em}\hspace{0.17em}}{x}^2+5x=p$

$\begin{array}{l}p(p+6)=72\\ p^2+6p-72=0\\ p^2+12p-6p-72=0\\ p(p+12)-6(p+12)=0\\ (p+12)(p-6)=0\end{array}$

$\begin{array}{l}p=6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}p=-12\\ x^2+5x-6=0,x^2+5x+12=0\\ x=\frac{-5\pm \sqrt{-23}}{2}\\ x=1\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}x=-6\\ x=1\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}x=-6,x=\frac{-5+\sqrt{-23}}{2}\end{array}$