Answer
$x= -1$
Work Step by Step
$\dfrac{x+3}{x^2+5x+6} = \dfrac{3}{2x+4}-\dfrac{1}{x+3}$
$\dfrac{x+3}{(x+2)(x+3)} = \dfrac{3}{2(x+2)}-\dfrac{1}{x+3}$
$\dfrac{x+3}{(x+2)(x+3)} = \dfrac{3(x+3) - 2(x+2)}{2(x+2)(x+3)}$
$\dfrac{x+3}{(x+2)(x+3)} = \dfrac{3x+9 - 2x-4}{2(x+2)(x+3)}$
$\dfrac{x+3}{(x+2)(x+3)} = \dfrac{x+5}{2(x+2)(x+3)}$
$\dfrac{x+3}{(x+2)(x+3)} - \dfrac{x+5}{2(x+2)(x+3)} = 0$
$\dfrac{2(x+3)-(x+5)}{2(x+2)(x+3)} = 0$
$\dfrac{2x+6-x-5}{2(x+2)(x+3)} = 0$
$\dfrac{x+1}{2(x+2)(x+3)} = 0$
$x + 1 = 0$
$x= -1$
