Answer
$y=\pm \dfrac{4}{5}i$
Work Step by Step
Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the given equation, $
25y^2+16=0
,$ are
\begin{array}{l}\require{cancel}
25y^2=-16
\\\\
y^2=\dfrac{-16}{25}
\\\\
y=\pm\sqrt{\dfrac{-16}{25}}
\\\\
y=\pm\sqrt{-1}\cdot\sqrt{\dfrac{16}{25}}
\\\\
y=\pm i\cdot\sqrt{\left( \dfrac{4}{5} \right)^2}
\\\\
y=\pm i\cdot\dfrac{4}{5}
\\\\
y=\pm \dfrac{4}{5}i
.\end{array}
