Answer
$z=\left\{ -3,\dfrac{13}{4} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
4z^2-z=39
,$ use first the properties of equality to express the equation in the form $x^2\pm bx=c.$ Once in this form, complete the square by adding $\left( \dfrac{b}{2} \right)^2$ to both sides of the equal sign. Then express the left side as a square of a binomial while simplify the right side. Then take the square root of both sides (Square Root Property) and use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{4z^2-z}{4}=\dfrac{39}{4}
\\\\
z^2-\dfrac{1}{4}z=\dfrac{39}{4}
.\end{array}
In the equation above, $b=
-\dfrac{1}{4}
.$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to
\begin{array}{l}\require{cancel}
\left( \dfrac{-\dfrac{1}{4}}{2} \right)^2
\\\\=
\left( -\dfrac{1}{4}\div2 \right)^2
\\\\=
\left( -\dfrac{1}{4}\cdot\dfrac{1}{2} \right)^2
\\\\=
\left( -\dfrac{1}{8} \right)^2
\\\\=
\dfrac{1}{64}
.\end{array}
Adding the value of $\left( \dfrac{b}{2} \right)^2,$ to both sides of the equation above results to
\begin{array}{l}\require{cancel}
z^2-\dfrac{1}{4}z+\dfrac{1}{64}=\dfrac{39}{4}+\dfrac{1}{64}
\\\\
z^2-\dfrac{1}{4}z+\dfrac{1}{64}=\dfrac{624}{64}+\dfrac{1}{64}
\\\\
z^2-\dfrac{1}{4}z+\dfrac{1}{64}=\dfrac{625}{64}
.\end{array}
With the left side now a perfect square trinomial, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( z-\dfrac{1}{8} \right)^2=\dfrac{625}{64}
.\end{array}
Taking the square root of both sides (Square Root Property), simplifying the radical and then isolating the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
z-\dfrac{1}{8}=\pm\sqrt{\dfrac{625}{64}}
\\\\
z-\dfrac{1}{8}=\pm\dfrac{25}{8}
\\\\
z=\dfrac{1}{8}\pm\dfrac{25}{8}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
z=\dfrac{1}{8}-\dfrac{25}{8}
\\\\
z=-\dfrac{24}{8}
\\\\
z=-3
\\\\\text{OR}\\\\
z=\dfrac{1}{8}+\dfrac{25}{8}
\\\\
z=\dfrac{26}{8}
\\\\
z=\dfrac{13}{4}
.\end{array}
Hence, $
z=\left\{ -3,\dfrac{13}{4} \right\}
.$