[tex]\bf \textit{volume of a pyramid}\\\\
V=\cfrac{1}{3}Bh\qquad
\begin{cases}
B=base=length\times width\\\\
h=height\\
----------\\
\textit{for a square pyramid}\\
\textit{the base is a square}\\
\textit{thus length=width=x}\\
B=x\cdot x=x^2\\
----------\\
V=141\frac{2}{3}\\\\
h=3x+2
\end{cases}[/tex]
[tex]\bf V=\cfrac{1}{3}Bh\implies 141\frac{2}{3}=x^2(3x+2)
\\\\\\
\cfrac{425}{3}=3x^3+2x^2\implies 425=9x^3+6x^2\implies 0=9x^3+6x^2-425[/tex]
solve for "x"
hmm I tried factoring it, tried a few combos, no dice, haven't graphed it
but, it doesn't seem it factor that neatly, so the solutions are likely decimals with long floats
