#### Answer

$x=-1$

#### Work Step by Step

Isolate the radical on the right side to obtain:
$3x=\sqrt{3x+7} -5
\\3x+5 = \sqrt{3x+7}$
Square both sides of the equation to obtain:
$(3x+5)^2=3x+7$
Use the rule $(a+b)^2 = a^2 + 2ab+b^2$ where $a=3x$ and $b=5$ to obtain:
$\\(3x)^2+2(3x)(5)+5^2=3x+7
\\9x^2+30x+25=3x+7$
Move all terms to the left side of the equation. Note that when a term is moved to the other side of the equation, its sign changes to its opposite.
$9x^2+30x+25-3x-7=0
\\9x^2+27x+18=0$
Factor out $9$ to obtain:
$9(x^2+3x+2)=0$
Divide 9 on both sides to obtain:
$x^2+3x+2=0$
Factor the trinomial to obtain:
$(x+2)(x+1)=0$
Equate each factor to zero then solve each equation to obtain:
$x+2 = 0 \text{ or } x+1 = 0
\\x = -2 \text{ or } x = -1$
Note that when $x=-2$,
$3(-2) - \sqrt{3(-2) + 7}\stackrel{?}{=}-5
\\-6 - \sqrt{-6+7}\stackrel{?}{=}-5
\\-6 - \sqrt{1} \stackrel{?}{=}-5
\\-6-1 \stackrel{?}{=}-5
\\-7 \ne -5$
Thus, $-2$ is an extraneous solution.
Therefore he only solution is $x=-1$.