Algebra and Trigonometry 10th Edition

$\frac{-3+\sqrt {21}}{6}, \frac{-3-\sqrt {21}}{6}$
$1/x - 1/(x+1)=3$ $x+1 - x= 3x(x+1)$ $1 = 3x^2+3x$ $3x^2+3x-1=0$ Using the discriminant rule for quadratic equations to find the roots we get x= $\frac{-3+\sqrt {21}}{6}$ or $\frac{-3-\sqrt {21}}{6}$