## Algebra: A Combined Approach (4th Edition)

$x=\dfrac{3}{2}\pm\dfrac{\sqrt{29}}{2}$
$\dfrac{x^{2}}{3}-x=\dfrac{5}{3}$ Multiply the whole equation by $3$ to avoid working with fractions: $3\Big(\dfrac{x^{2}}{3}-x=\dfrac{5}{3}\Big)$ $x^{2}-3x=5$ Take the $5$ to the left side of the equation: $x^{2}-3x-5=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here $a=1$, $b=-3$ and $c=-5$ Substitute: $x=\dfrac{-(-3)\pm\sqrt{(-3)^{2}-4(1)(-5)}}{2(1)}=\dfrac{3\pm\sqrt{9+20}}{2}=...$ $...=\dfrac{3\pm\sqrt{29}}{2}=\dfrac{3}{2}\pm\dfrac{\sqrt{29}}{2}$