## Calculus (3rd Edition)

$$\frac{10}{7}$$
Since $$f'(x)= 2x + 3$$ Then at $x=2,$ $m= f'(2)=7$. Hence, the tangent line is \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-10}{x-2}&=7\\ y&=7x-4 \end{align*} Since the tangent line intersect with the $x-$ axis at $x=4/7$, then $Q$ has coordinates $( 4/ 7, 0)$, $R$ has coordinates $(2, 0)$, and the subtangent is $$2-\frac{4}{7}=\frac{10}{7}$$