Answer
$2in,2in, 5in$
Work Step by Step
Let $x$ be the length of the base's side. Then our equation for the volume is: $x^2(x+3)=20\\x^3+3x^2-20=0\\(x-2)(x^2+5x+10)=0.$
$x^2+5x+10$ cannot be $0$ because its determinant is $0$, thus the only solution of the equation is $x=2$. Thus the dimensions are $2in,2in, 5in$
