Answer
$S=\dfrac{t\pm5|t|}{12}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
S(6S-t)=t^2
,$ is equivalent to
\begin{align*}\require{cancel}
6S^2-tS&=t^2
\\
6S^2-tS-t^2&=0
.\end{align*}
The equation above has
\begin{align*}
a=
6
,\text{ }b=
-t
,\text{ and }c=
-t^2
.\end{align*}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
S&=
\dfrac{-(-t)\pm\sqrt{(-t)^2-4(6)(-t^2)}}{2(6)}
\\\\&=
\dfrac{t\pm\sqrt{t^2+24t^2}}{12}
\\\\&=
\dfrac{t\pm\sqrt{25t^2}}{12}
\\\\&=
\dfrac{t\pm\sqrt{25}\cdot\sqrt{t^2}}{12}
\\\\&=
\dfrac{t\pm5|t|}{12}
&(\text{use }\sqrt{x^2}=|x|)
.\end{align*}
Hence, in terms of $S,$ the given equation is equivalent to $S=\dfrac{t\pm5|t|}{12}$.
