[tex]\textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=h-3\\ V=25h\pi \end{cases}\implies 25h\pi =\pi (h-3)^2 h \\\\\\ \cfrac{25h\pi }{\pi h }=(h-3)(h-3)h\implies 25=\stackrel{F~O~I~L}{h^2-6h+9}\implies 0=h^2-6h-16 \\\\\\ 0=(h-8)(h+2)\implies \blacktriangleright h= \begin{cases} 8~~\checkmark\\ -2 \end{cases} \blacktriangleleft ~\hfill \blacktriangleright \underset{radius}{\stackrel{8~~ - ~~3}{5}} \blacktriangleleft[/tex]
notice, we didn't use h = -2, because the height cannot be a negative value.
