Answer
$x=-4$ or $x=1$
Work Step by Step
We solve by factoring:
$x^{2}+3x+\sqrt{x^{2}+3x}=6$
$x^{2}+3x+\sqrt{x^{2}+3x}-6=0$
$[\sqrt{x^2+3x}+3][\sqrt{x^2+3x}-2]=0$
$\sqrt{x^2+3x}+3=0$ or $\sqrt{x^2+3x}-2=0$
$\sqrt{x^2+3x}=-3$ is not possible because a square root is never negative.
So: $\sqrt{x^2+3x}=2$
$x^2+3x=(2)^2$
$x^2+3x=4$
$x^2+3x-4=0$
$(x+4)(x-1)=0$
$x=-4$ or $x=1$