Answer
$y=\dfrac{3\pm\sqrt{17}}{4}$
Work Step by Step
Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), then the solutions to the given equation, $
\left( y+\dfrac{3}{4} \right)^2=\dfrac{17}{16}
,$ are
\begin{array}{l}\require{cancel}
y+\dfrac{3}{4}=\pm\sqrt{\dfrac{17}{16}}
\\\\
y+\dfrac{3}{4}=\pm\dfrac{\sqrt{17}}{\sqrt{16}}
\\\\
y+\dfrac{3}{4}=\pm\dfrac{\sqrt{17}}{\sqrt{(4)^2}}
\\\\
y+\dfrac{3}{4}=\pm\dfrac{\sqrt{17}}{4}
\\\\
y=\dfrac{-3}{4}\pm\dfrac{\sqrt{17}}{4}
\\\\
y=\dfrac{-3\pm\sqrt{17}}{4}
.\end{array}
