Answer
$x_{1}=0$
$x_{2}=-2$
Work Step by Step
Factor $(x^2+2x)^{1/2}$:
$(x^2+2x)^{1/2}(3x-2(x^2+2x))=0$
$(x^2+2x)^{1/2}(3x-2x^2-4x)=0$
$(x^2+2x)^{1/2}(-2x^2-x)=0$
First part:
$(x^2+2x)^{1/2}=0$
$x^2+2x=0$
$x(x+2)=0$
$x_{1}=0$
$x_{2}=-2$
Second part:
$(-2x^2-x)=0$
$-x(2x+1)=0$
$2x+1=0$
$x_3=-\frac{1}{2}$
Looking at the solutions, the first and second ones are real solutions, but the third one is not because it will result in a negative value inside the radicals.
