Answer
$10.83$ hours or $12.50$ hours
Work Step by Step
Recardo wants to know the time it will take him to be $50$ miles away from Fresno. Substitute $50$ in place of $D(t)$ and solve for $t$.
$$\begin{aligned}
D(t) &= \lvert 700-60t\rvert\\
50 &= \lvert 700-60t\rvert.
\end{aligned}$$ Rewrite into two equations and solve. $$\begin{aligned}
700-60t&= 50\\
-60t&=50-700\\
-60t& = -650\\
60t& = 650\\
t&= \frac{650}{60}\\
&= \frac{65}{6}\\
&\approx 10.83
\end{aligned}$$ or $$\begin{aligned}
700-60t&= -50\\
-60t&=-50-700\\
-60t& = -750\\
60t& = 750\\
t&= \frac{750}{60}\\
&= \frac{75}{6}\\
&= 12.50.
\end{aligned}$$ Check the solution. $$\begin{aligned}
D(65/6) &= \lvert 700-60\cdot \frac{65}{6}\rvert\\\\
& = \lvert 700-10\cdot 65\rvert\\\\
& = \lvert 700-650\rvert\\
& = \lvert 50\rvert\\
& = 50
\end{aligned}$$ and $$\begin{aligned}
D(65/6) &= \lvert 700-60\cdot \frac{75}{6}\rvert\\\\
& = \lvert 700-10\cdot 75\rvert\\\\
& = \lvert 700-750\rvert\\
& = \lvert -50\rvert\\
& = 50.
\end{aligned}$$ Hence, Ricardo will be $50$ miles away from Fresno at either $10.83$ hours later of at $12.50$ hours later.
