Answer
$x=-2$
Work Step by Step
We are given:
$\sqrt{x+2}=1-\sqrt{3x+7}$
We square both sides:
$(\sqrt{x+2})^{2}=(1-\sqrt{3x+7})^{2}$
$(\sqrt{x+2})(\sqrt{x+2})=(1-\sqrt{3x+7})(1-\sqrt{3x+7})$
$x+2=1-2\sqrt{3x+7}+(3x+7)$
$x+2-3x-8=-2\sqrt{3x+7}$
$-2x-6=-2\sqrt{3x+7}$
$2\sqrt{3x+7}=2x+6$
$2\sqrt{3x+7}=2(x+3)$
$\sqrt{3x+7}=x+3$
We square both sides again:
$(\sqrt{3x+7})^{2}=(x+3)^{2}$
$3x+7=x^{2}+6x+9$
$x^{2}+3x+2=0$
And factor:
$(x+2)(x+1)=0$
Use the zero-factor property by equating each factor to zero:
$(x+2)=0$ or $(x+1)=0$
$x=-2$ or $x=-1$
However, the solution $x=-1$ does not work in the original equation:
$\sqrt{-1+2}=1-\sqrt{3*-1+7}$
$\sqrt{1}=1-\sqrt{-3+7}$
$1=1-\sqrt{4}$
$1=1-2$
$1=-1$
Since we got a false statement, the solution $x=-1$ does not work. Thus the only solution is
$x=-2$.