## Intermediate Algebra: Connecting Concepts through Application

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