Prove $\lim_{x\to 2} \sqrt{x^2+5} = 3$ As stated in the title, I need to prove that $\lim_{x\to 2} \sqrt{x^2+5} = 3$ using only the precise definition of a limit.

For any given $\varepsilon \gt 0$, there exists a $\delta = $
Such that $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} \rvert \lt \varepsilon $
I've attempted to convert $\sqrt{x^2+5}$ into the following:
$\sqrt{x^2+5} = \sqrt{(x^2-4)+9}\\
 \sqrt{x^2+5} = \sqrt{(x^2-4x+4)+4x +1} = \sqrt{(x-2)^2+4(x-2)+9} $
I have a hunch that I am heading in the wrong direction. Can I get some advice on what I might be doing wrong?

We should prove that,for any given $ε>0$, there exists a $δ$ if $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} -3\rvert \lt \varepsilon$
$$\left| \sqrt { { x }^{ 2 }+5 } -3 \right| =\left| \frac { { x }^{ 2 }-4 }{ \sqrt { { x }^{ 2 }+5 } +3 }  \right| =\left| \frac { { \left( x-2 \right)  }\left( x-2+4 \right)  }{ \sqrt { { x }^{ 2 }+5 } +3 }  \right| <\frac { \left| { \left( x-2 \right)  }^{ 2 }+4\left( x-2 \right)  \right|  }{ 3 } <\frac { { \left| x-2 \right|  }^{ 2 }+4\left| x-2 \right|  }{ 3 } <\frac { { \delta  }^{ 2 }+4\delta  }{ 3 } =\epsilon $$
