Answer
$t=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
At^2+Bt=-C
,$ is equivalent to
\begin{align*}\require{cancel}
At^2+Bt+C&=0
.\end{align*}
The equation above has
\begin{align*}
a=
A
,\text{ }b=
B
,\text{ and }c=
C
.\end{align*}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
t&=
\dfrac{-B\pm\sqrt{B^2-4(A)(C)}}{2(A)}
\\\\&=
\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}
.\end{align*}
Hence, in terms of $t,$ the given equation is equivalent to $t=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}$.
