Answer
$$x=-1, -\frac{1}{4}$$
Work Step by Step
$$\frac{4x+3}{x+1}+\frac{2}{x}=\frac{1}{x^2+x}$$
$$(x(x+1))(\frac{4x+3}{x+1}+\frac{2}{x})=(x(x+1))(\frac{1}{x^2+x})$$
$$x(4x+3)+2(x+1)=1$$
$$4x^2+3x+2x+2=1$$
$$4x^2+5x+1=0$$
$$x=\frac{-5+\sqrt (25-16)}{8}$$
$$x=\frac{-5+3}{8}$$
$$x=\frac{-1}{4}$$
$$x=\frac{-5-\sqrt (25-16)}{8}$$
$$x=\frac{-8}{8}$$
$$x=-1$$
$$x=-1, -\frac{1}{4}$$
