Answer
$\left\{\dfrac{5}{4}, \dfrac{2}{3}\right\}$
Work Step by Step
Write the equation in the form $ax^2+bx+c=0$ to obtain:
\begin{align*}
12t^2-23t+40-30&=0\\
12t^2-23t+10&=0
\end{align*}
Solve the equation using the $\text{ac-method}$.
With $a=12$ and $c=10$, $ac= 12(10)=120$.
Recall:
$ax^2+bx+c$ can be factored if there exists $d$ and $e$ such that $ac=de$ and $d+e=b$.
The trinomial can then be factored by grouping by rewriting the trinomial as $ax^2+dx+ex+c$.
Look for factors of $120$ whose sum is $-23$.
Note that $120=-15(-8)$ and $-15+(-8)=-23$.
This means that the trinomial is factorable and $d=-15$ and $e=-8$.
Solve the given equation by grouping:
\begin{align*}
12t^2-23t+10&=0\\
12t^2-15t-8t+10&=0\\
(12t^2-15t)+(-8t+10)&=0\\
3t(4t-5)+(-2)(4t-5)&=0\\
(4t-5)(3t-2)&=0\
\end{align*}
Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain:
\begin{align*}
4t-5&=0 &\text{or}& &3t-2=0\\
4t&=5 &\text{or}& &3t=2\\
t&=\frac{5}{4} &\text{or}& &t=\frac{2}{3}
\end{align*}
