read the section on completing the square
Pauls Online Notes : Algebra - Quadratic Equations - Part II
Hello, smyatsallie!
$\displaystyle \begin{array}{c} 8y^2 - 22y + 15 \;=\;0 \\ \\ y^2 - \frac{11}{4}y + \frac{15}{8} \;=\;0 \\ \\ y^2 - \frac{11}{4}y \;=\;-\frac{15}{8} \\ \\ y^2 - \frac{11}{4}y + \frac{121}{64} \;=\;-\frac{15}{8} + \frac{121}{64} \\ \\ \left(y - \frac{11}{8}\right)^2 \;=\;\frac{1}{64} \end{array}$
Take the square root of both sides:
. . $\displaystyle y - \tfrac{11}{8} \;=\;\pm\sqrt{\tfrac{1}{64}}$
. . $\displaystyle y - \tfrac{11}{8} \;=\;\pm\tfrac{1}{8} \quad\Rightarrow\quad y \;=\;\tfrac{11}{8} \pm \tfrac{1}{8} $
Therefore: .$\displaystyle \begin{Bmatrix}y &=& \frac{11}{8} + \frac{1}{8} &=& \frac{12}{8} &=& \frac{3}{2} \\ \\ y &=& \frac{11}{8} - \frac{1}{8} &=& \frac{10}{8} &=& \frac{5}{4} \end{Bmatrix}$