Answer
$\left\{ 1-\sqrt{3},1+\sqrt{3} \right\}$
Work Step by Step
Multiplying both sides of the given equation, $
\dfrac{6}{x^2}=\dfrac{3}{x+1}
,$ by the $LCD=
x^2(x+1)
$, then,
\begin{array}{l}\require{cancel}
x^2(x+1)\left( \dfrac{6}{x^2} \right)=\left( \dfrac{3}{x+1} \right)x^2(x+1)
\\\\
(x+1)(6)=3(x^2)
\\\\
6x+6=3x^2
\\\\
-3x^2+6x+6=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
-3x^2+6x+6=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(6)\pm\sqrt{(6)^2-4(-3)(6)}}{2(-3)}
\\\\=
\dfrac{-6\pm\sqrt{36+72}}{-6}
\\\\=
\dfrac{-6\pm\sqrt{108}}{-6}
\\\\=
\dfrac{-6\pm\sqrt{36\cdot3}}{-6}
\\\\=
\dfrac{-6\pm\sqrt{(6)^2\cdot3}}{-6}
\\\\=
\dfrac{-6\pm6\sqrt{3}}{-6}
\\\\=
\dfrac{-6(1\pm\sqrt{3})}{-6}
\\\\=
\dfrac{\cancel{-6}(1\pm\sqrt{3})}{\cancel{-6}}
\\\\=
1\pm\sqrt{3}
.\end{array}
Upon checking, both solutions satisfy the original equation. Hence, the solutions are $
\left\{ 1-\sqrt{3},1+\sqrt{3} \right\}
.$
