Finding the real solutions of the radical equation $\sqrt{3x+10}-\sqrt{x+2}=2$

Question

Find the solutions to the following:$$\sqrt{3x+10}-\sqrt{x+2}=2$$

This is what I tried so far: \begin{align*} (\sqrt{3x+10}-\sqrt{x+2})^2 & =2\\ (3x+10)+(x+2)-2\sqrt{(x+2)(3x+10)} & =2\\ 2x+5-\sqrt{3x^2+16x+20} & =0 \end{align*}

Now I do not know where to go from here...

Answer 1

Hint:

$2x+5-\sqrt{3x^2+16x+20}=0\iff2x+5=\sqrt{3x^2+16x+20} \Longrightarrow\\(2x+5)^2=3x^2+16x+20.$

Answer 2

Rather than squaring both sides as stated, it might be more useful to use $$ \sqrt{3x+10}=2+\sqrt{x+2} $$ which can then be squared to get $$ 3x+10 = 6+x+4\sqrt{x+2} $$ You can probably finish it from here for yourself.