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

$n=\dfrac{9}{22}\pm\dfrac{5\sqrt{5}}{22}$
$11n^{2}-9n=1$ Take the $1$ to the left side of the equation: $11nY{2}-9n-1=0$ Use the quadratic formula to solve this equation. The formula is $n=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=11$, $b=-9$ and $c=-1$: Substitute: $n=\dfrac{-(-9)\pm\sqrt{(-9)^{2}-4(11)(-1)}}{2(11)}=\dfrac{9\pm\sqrt{81+44}}{22}=...$ $...=\dfrac{9\pm\sqrt{125}}{22}=\dfrac{9\pm5\sqrt{5}}{22}=\dfrac{9}{22}\pm\dfrac{5\sqrt{5}}{22}$