Solve the radical equation $ x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$ 
Solve the following equation:
  $$ x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$$

I wanted to solve this equation. First I tried to change the equations under the roots to the complete square to simplify them out, but it just became more complicated.
Can someone help me with this equation, please?

Assuming you want real roots, you can use this trick . . .

Let $f\colon \mathbb{R} \to \mathbb{R}$ be given by $f(t) = t\sqrt{t^2+5}$.

Then $f$ is an odd function.

Also, $f$ is strictly increasing, hence $f$ is one-to-one.

Then, letting $u=2x+1$,
\begin{align*}
&x\sqrt{x^2+5}+(2x+1)\sqrt{4x^2 + 4x + 6}=0
\qquad\qquad\qquad\qquad\;\;
\\[4pt]
\iff\;&x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt]
\iff\;&f(x) + f(u)=0\\[4pt]
\iff\;&f(x) = -f(u)\\[4pt]
\iff\;&f(x) = f(-u)\qquad\text{[since $f$ is odd]}\\[4pt]
\iff\;&x = -u\qquad\qquad\;\;\text{[since $f$ is one-to-one]}\\[4pt]
\iff\;&x = -(2x+1)\\[4pt]
\iff\;&3x+1 = 0\\[4pt]
\iff\;&x = -{\small{\frac{1}{3}}}\\[4pt]
\end{align*}
As an alternative, using the same trick, you can get a factored form:
\begin{align*}
&x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt]
\implies\;&x\sqrt{x^2+5}=-u\sqrt{u^2 + 5}\\[4pt]
\implies\;&x^2(x^2+5)=u^2(u^2 + 5)\\[4pt]
\implies\;&x^4+5x^2=u^4+5u^2\\[4pt]
\implies\;&(u^4-x^4)+5(u^2-x^2)=0\\[4pt]
\implies\;&(u^2-x^2)(u^2+x^2)+5(u^2-x^2)=0\\[4pt]
\implies\;&(u^2-x^2)(u^2+x^2+5)=0\\[4pt]
\implies\;&(u-x)(u+x)(u^2+x^2+5)=0\\[4pt]
\implies\;&(x+1)(3x+1)(5x^2+4x+6)=0\qquad\text{[replacing $u$ by $2x+1$]}
\end{align*}
The two candidate real roots, $x=-1,\;x=-{\large{\frac{1}{3}}}\;$need to be verified against the original equation since, when squaring both sides, extraneous real roots were potentially introduced. In this case, as it turns out, the candidate root $x=-{\large{\frac{1}{3}}}$ is ok, but the candidate root $x=-1$ fails, so is not an actual root.
