Answer
$v\leq60$.
Work Step by Step
$d=v+\frac{v^2}{20}$.
for $d\lt240$.
$v+\frac{v^2}{20}\leq240$, multiplying both sides by $20$.
$20v+v^2\leq4800$,
$v^2+20v-4800\leq0$,
Solving for the trinomial using the quadratic formula for the quadratic equation of $ax^2+bx+c, x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
In this case, $v^2+20v-4800\leq0$,
$v=\frac{-20\pm\sqrt{20^2-4(1)(-4800)}}{2(1)}=\frac{-20\pm140}{2}$
thus, $v=-80$ or $v=60$. Since we're solving for speed, only positive answers are relevant. Therefore, $v\leq60$.
