#### Answer

$x=1\pm\sqrt{3}$

#### Work Step by Step

$\dfrac{6}{x^{2}}=\dfrac{3}{x+1}$
Multiply the whole equation by $x^{2}(x+1)$
$x^{2}(x+1)\Big(\dfrac{6}{x^{2}}=\dfrac{3}{x+1}\Big)$
$6(x+1)=3x^{2}$
Evaluate the product on the left side:
$6x+6=3x^{2}$
Take all terms to the right side of the equation and simplify it by taking out common factor $3$:
$0=3x^{2}-6x-6$
$3(x^{2}-2x-2)=0$
Take the $3$ to divide the right side of the equation:
$x^{2}-2x-2=\dfrac{0}{3}$
$x^{2}-2x-2=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=1$, $b=-2$ and $c=-2$.
Substitute the known values into the formula and simplify:
$x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-2)}}{2(1)}=\dfrac{2\pm\sqrt{4+8}}{2}=...$
$...=\dfrac{2\pm\sqrt{12}}{2}=\dfrac{2\pm2\sqrt{3}}{2}=1\pm\sqrt{3}$
The original equation is not undefined for neither of the solutions found. Our final answers are $x=1\pm\sqrt{3}$