Answer
$z=\dfrac{-3\pm\sqrt{5}}{2}$
Work Step by Step
Using the completing the square method, the solutions of the given quadratic equation, $
z^2+3z+1=0
,$ are
\begin{array}{l}\require{cancel}
z^2+3z=-1
\\\\
z^2+3z+\left( \dfrac{3}{2} \right)^2=-1+\left( \dfrac{3}{2} \right)^2
\\\\
z^2+3z+\dfrac{9}{4}=-1+\dfrac{9}{4}
\\\\
\left( z+\dfrac{3}{2} \right)^2=-\dfrac{4}{4}+\dfrac{9}{4}
\\\\
\left( z+\dfrac{3}{2} \right)^2=\dfrac{5}{4}
\\\\
z+\dfrac{3}{2}=\pm\sqrt{\dfrac{5}{4}}
\\\\
z+\dfrac{3}{2}=\pm\dfrac{\sqrt{5}}{\sqrt{4}}
\\\\
z+\dfrac{3}{2}=\pm\dfrac{\sqrt{5}}{2}
\\\\
z=-\dfrac{3}{2}\pm\dfrac{\sqrt{5}}{2}
\\\\
z=\dfrac{-3\pm\sqrt{5}}{2}
.\end{array}
