Answer
$z=3$
Work Step by Step
Squaring both sides, the given equation, $
z=\sqrt{\dfrac{5z+3}{2}}
,$ is equivalent to
\begin{align*}
(z)^2&=\left(\sqrt{\dfrac{5z+3}{2}}\right)^2
\\\\
z^2&=\dfrac{5z+3}{2}
.\end{align*}
In the form $ax^2+bx+c=0,$ the equation abovec is equivalent to
\begin{align*}\require{cancel}
2(z^2)&=\left(\dfrac{5z+3}{\cancel2}\right)\cancel2
\\\\
2z^2&=5z+3
\\
2z^2-5z-3&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-3)(2z+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-3=0 & 2z+1=0
\\
z=3 & 2z=-1
\\\\
& z=-\dfrac{1}{2}
.\end{array}
Checking by substituting the solutions in the given equation results to
\begin{array}{l|r}
\text{If }z=3: & \text{If }z=-\dfrac{1}{2}:
\\\\
3\overset{?}=\sqrt{\dfrac{5(3)+3}{2}} &
-\dfrac{1}{2}\overset{?}=\sqrt{\dfrac{5\left(-\dfrac{1}{2}\right)+3}{2}}
\\\\
3\overset{?}=\sqrt{\dfrac{15+3}{2}} &
-\dfrac{1}{2}\ne\text{ some nonnegative number}.
\\\\
3\overset{?}=\sqrt{\dfrac{18}{2}} &
\\
3\overset{?}=\sqrt{9} &
\\
3\overset{\checkmark}=3 &
\end{array}
Since $z=-\dfrac{1}{2}$ does not satisfy the original equation then the only solution of the equation $
z=\sqrt{\dfrac{5z+3}{2}}
$ is $z=3$.
