#### Answer

$\left\{\dfrac{9}{2}, -\dfrac{2}{5}\right\}$

#### Work Step by Step

Write the equation in the form $ax^2+bx+c=0$ to obtain:
$$10t^2-41t-18=0$$
Solve the equation using the $\text{ac-method}$.
With $a=10$ and $c=-18$, $ac= 10(-18)=-180$.
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 tfactored by grouping by rewriting the trinomial as $ax^2+dx+ex+c$.
Look for factors of $-180$ whose sum is $-41$.
Note that $-180=-45(4)$ and $-45+4=-41$.
This means that the trinomial is factorable.
Solve the given equation by grouping:
\begin{align*}
10t^2-41t-18&=0\\
10t^2-45t+4t-18&=0\\
(10t^2-45t)+(4t-18)&=0\\
5t(2t-9)+2(2t-9)&=0\\
(2t-9)(5t+2)&=0\end{align*}
Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain:
\begin{align*}
2t-9&=0 &\text{or}& &5t+2=0\\
2t&=9 &\text{or}& &5t=-2\\
t&=\frac{9}{2} &\text{or}& &t=-\frac{2}{5}\\
\end{align*}