Answer
$w=\left\{ -\dfrac{8}{3},3 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
3w^2-w=24
,$ 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{3w^2-w}{3}=\dfrac{24}{3}
\\\\
w^2-\dfrac{1}{3}w=8
.\end{array}
In the equation above, $b=
-\dfrac{1}{3}
.$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to
\begin{array}{l}\require{cancel}
\left( \dfrac{-\dfrac{1}{3}}{2} \right)^2
\\\\=
\left( -\dfrac{1}{3}\div2 \right)^2
\\\\=
\left( -\dfrac{1}{3}\cdot\dfrac{1}{2} \right)^2
\\\\=
\left( -\dfrac{1}{6} \right)^2
\\\\=
\dfrac{1}{36}
.\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}
w^2-\dfrac{1}{3}w+\dfrac{1}{36}=8+\dfrac{1}{36}
\\\\
w^2-\dfrac{1}{3}w+\dfrac{1}{36}=\dfrac{288}{36}+\dfrac{1}{36}
\\\\
w^2-\dfrac{1}{3}w+\dfrac{1}{36}=\dfrac{289}{36}
.\end{array}
With the left side now a perfect square trinomial, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( w-\dfrac{1}{6} \right)^2=\dfrac{289}{36}
.\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}
w-\dfrac{1}{6}=\pm\sqrt{\dfrac{289}{36}}
\\\\
w-\dfrac{1}{6}=\pm\dfrac{17}{6}
\\\\
w=\dfrac{1}{6}\pm\dfrac{17}{6}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
w=\dfrac{1}{6}-\dfrac{17}{6}
\\\\
w=-\dfrac{16}{6}
\\\\
w=-\dfrac{8}{3}
\\\\\text{OR}\\\\
w=\dfrac{1}{6}+\dfrac{17}{6}
\\\\
w=\dfrac{18}{6}
\\\\
w=3
.\end{array}
Hence, $
w=\left\{ -\dfrac{8}{3},3 \right\}
.$
