Answer
The solutions are $x=-3, x=\dfrac{1+\sqrt{13}}{2}$ and $x=\dfrac{1-\sqrt{13}}{2}$
Work Step by Step
$x^2\sqrt{x+3}=(x+3)^{3/2}\hspace{0.7cm}{\color{blue}{\text{Given equation}}}$
$\Rightarrow x^2\sqrt{x+3}-(x+3)^{3/2}=0$
$\Rightarrow \sqrt{x+3}(x^2-(x+3))=0$
$\Rightarrow \sqrt{x+3}(x^2-x-3)=0$
$\sqrt{x+3}=0\Rightarrow x+3=0\Rightarrow x=-3$
$\text{or }x^2-x-3=0$
Use the quadratic formula where $a=1, b=-1$ and $c=-3$
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-3)}}{2(1)}$
$\Rightarrow x=\dfrac{1\pm\sqrt{1+12}}{2}=\dfrac{1\pm\sqrt{13}}{2}$
The solutions are $x=-3, x=\dfrac{1+\sqrt{13}}{2}$ and $x=\dfrac{1-\sqrt{13}}{2}$
