## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$$x=\frac{\sqrt{61}-1}{6},\:x=\frac{-\sqrt{61}-1}{6}$$
To find the x-intercepts, we set the equation equal to zero and then complete the square. Doing this, we find: $$3x^2+x=5\\ x^2+\frac{x}{3}=\frac{5}{3}\\ x^2+\frac{x}{3}+\left(\frac{1}{6}\right)^2=\frac{5}{3}+\left(\frac{1}{6}\right)^2\\ \left(x+\frac{1}{6}\right)^2=\frac{61}{36}\\ x=\frac{\sqrt{61}-1}{6},\:x=\frac{-\sqrt{61}-1}{6}$$