It is given that $\sqrt{3p^3 + 6r^2} = \frac{5r}{2}.$ Make r the subject of the equation.

1. Remove the square root by squaring both sides.

$\sqrt{3p^3+6r^2}=\frac{5r}{2}$
$3p^3+6r^2=\left(\frac{5r}{2}\right)^2$
$3p^3+6r^2=\frac{25r^2}{4}$

2. Clear the fraction and collect the $r^2$ terms.

$4(3p^3+6r^2)=25r^2$
$12p^3+24r^2=25r^2$
$12p^3=r^2$

3. Take the square root and choose the valid sign.

$r=\sqrt{12p^3}$
Since $\sqrt{3p^3+6r^2}\geq 0$, the original equation gives $\frac{5r}{2}\geq 0$, so $r\geq 0$. Hence the negative root is rejected.

$r=\sqrt{4p^2\cdot 3p}=2p\sqrt{3p}$