jmc

We can write the given equation as $$x^2+4x\sqrt{x+3}+4(x+3)=25.$$Then $$(x+2\sqrt{x+3}{)}^2=25,$$so $x+2\sqrt{x+3}=\pm 5.$ Then $$-x\pm 5=2\sqrt{x+3}.$$Squaring both sides, we get $x^2\pm 10x+25=4x+12.$

In the $+$ case, we get $$x^2+6x+13=0,$$which has no real solutions.

In the $-$ case, we get $$x^2-14x+13=0,$$which leads to the solutions 1 and 13. We check that only $\boxed{1}$ works.