Answer
$t=\dfrac{-3\pm\sqrt{14}}{2}$
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( t+\dfrac{3}{2} \right)^2=\dfrac{7}{2}
,$ are
\begin{array}{l}\require{cancel}
t+\dfrac{3}{2}=\pm\sqrt{\dfrac{7}{2}}
\\\\
t+\dfrac{3}{2}=\pm\sqrt{\dfrac{7}{2}\cdot\dfrac{2}{2}}
\\\\
t+\dfrac{3}{2}=\pm\sqrt{\dfrac{14}{4}}
\\\\
t+\dfrac{3}{2}=\pm\dfrac{\sqrt{14}}{\sqrt{4}}
\\\\
t+\dfrac{3}{2}=\pm\dfrac{\sqrt{14}}{(2)^2}
\\\\
t+\dfrac{3}{2}=\pm\dfrac{\sqrt{14}}{2}
\\\\
t=-\dfrac{3}{2}\pm\dfrac{\sqrt{14}}{2}
\\\\
t=\dfrac{-3\pm\sqrt{14}}{2}
.\end{array}