= Solution
The second-order remainder form of <Taylor's theorem> gives, for a point $\xi_x$ between $0$ and $x$,
$$
f(x)=-5+3x+\tfrac12f''(\xi_x)x^2.
$$
For $x\ne0$, the given curvature bounds imply
$$
-5+3x+x^2\leq f(x)<-5+3x+\tfrac32x^2.
$$
In particular $f(1)<-1/2<0$ and $f(2)\geq5>0$. Differentiability implies <continuity>, so the <intermediate value theorem> gives a root in $(1,2)$.
The <mean value theorem> applied to $f'$ gives $f'(x)=3+xf''(\eta_x)\geq3+2x>0$ for $x\in(1,2)$. Thus $f$ is strictly increasing there, and it cannot have two roots. \b[There is exactly one root in $(1,2)$.]
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