Okay, here we have this:
Considering the provided equation, we are going to solve it using the quadratic formula, so we obtain the following:
[tex]\begin{gathered} \mleft(x+3\mright)^2-5=0 \\ x^2+6x+9-5=0 \\ x^2+6x+4=0 \\ x_{1,\: 2}=\frac{-6\pm\sqrt{6^2-4\cdot\:1\cdot\:4}}{2\cdot\:1} \\ x_{1,\: 2}=\frac{-6\pm\:2\sqrt{5}}{2\cdot\:1} \\ x_1=\frac{-6+2\sqrt{5}}{2\cdot\:1},\: x_2=\frac{-6-2\sqrt{5}}{2\cdot\:1} \\ x=-3+\sqrt{5},\: x=-3-\sqrt{5} \end{gathered}[/tex]
Finally we obtain that the correct answer is the option A.
