The position of a particle moving along a line is given by $ 2t^3 -24t^2+90t + 7$ for $t >0$ For what values of $t$ is the speed of the particle increasing?
I tried to find the first derivative and I get
$$6t^2-48t+90 = 0$$
$$ t^2-8t+15 = 0$$
Which is giving me $ t>5$ and $0 < t < 3$, but the book gives a different answer

You need to keep in mind that speed is the absolute value of velocity.  If the velocity is positive and increasing, the speed will increasing.  But if the velocity is negative and decreasing (becoming more and more negative), the speed is increasing. 
There are two other cases to consider, but I'll leave that to you.
For your problem, let's consider the motion of the point by
looking at the velocity function (whose graph is a parabola opening upwards with zeros at $x=3$ and $x=5$). With the standard conventions:
From $t=0$ to $t=3$:  The particle is moving to the right. Its speed   is decreasing over $(0,3)$. (Note the velocity is positive and decreasing here.)
At $t=3$, the particle reverses direction. (The velocity is 0 here.)
From $t=3$ to $t=4$:  The particle is moving to the left. Its speed is increasing over $(3,4)$. (The velocity is negative and decreasing here.)
From $t=4$ to $t=5$:  The particle is moving to the left. Its speed is decreasing over $(4,5)$. (The velocity is negative and increasing here)
At $t=5$, the particle reverses direction. (The velocity is 0 here.)
From $t=5$ onwards:  The particle is moving to the right. Its speed is increasing over $(5,\infty)$. (The velocity is positive and increasing here)
