Answer
$$x=4$$
Work Step by Step
$$\sqrt{3x+4}+5=2x+1$$
$$\sqrt{3x+4}=2x-4$$
$$3x+4=4x^2-16x+16$$
$$0=4x^2-19x+12$$
$$0=(x-4)(4x-3)$$
$$x-4=0$$
$$x=4$$
$$4x-3=0$$
$$x=\frac{3}{4}$$
$\frac{3}{4}$ is an extraneous solution, so $x=4$.
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