The linear approximation to [tex]f(x)[/tex] centered at [tex]x=a[/tex] is
[tex]f(x)\approx f(a)+f'(a)(x-a)[/tex]
What this means is that you can use the tangent line to [tex]f(x)[/tex] at [tex]x=a[/tex] to get a decent approximation of [tex]f(x)[/tex] at some other value of [tex]x[/tex] to within a certain degree of accuracy depending on how close this value [tex]x[/tex] is close to [tex]a[/tex]. (Note that when [tex]x=a[/tex], the approximation is exact; [tex]f(a)=f(a)[/tex].)
So what you're asked to do is find an approximate value of a zero of [tex]f(x)[/tex] near [tex]a=3[/tex]. That is to say, you're looking for some value [tex]c[/tex] such that [tex]f(c)=0[/tex], but all you have at your disposal is the linear approximation to the function.
[tex]f(x)\approx f(3)+f'(3)(x-3)[/tex]
[tex]f(x)\approx2+5(x-2)[/tex]
[tex]f(x)\approx5x-8[/tex]
You know that if [tex]c[/tex] is a zero of [tex]f[/tex], then [tex]f(c)=0[/tex], so you get
[tex]f(c)=0\approx5c-8[/tex]
Solving for [tex]c[/tex], you find
[tex]0=5c-8\implies 5c=8\implies c=\dfrac85=1.6[/tex]
This means that an approximate zero of [tex]f(x)[/tex] is [tex]x=1.6[/tex].
