It is necessary to go from a point $A(0,0)$ to a point $B(a,b)$ walking from A to $P(x,0)$ with a speed $v_1$ and then until the point B with a speed $v_2$.
Find where is the point in which it is necessary to abandon x axis in order to have the minimum time to complete the path.
I've called $x$ the distance of A from P. The function is $$t(x)=\frac{x}{v_1}+\frac{\sqrt{(a-x)^2+b^2}}{v_2}$$ $$\frac{dt}{dx}=\frac{1}{v_1}+\frac{(a-x)*(-1)}{v_2* \sqrt{(a-x)^2+b^2}}=0 \Rightarrow v_2*\sqrt{(a-x)^2+b^2}=v_1*(a-x) \Rightarrow x^2(v_2^2-v_1^2)+x(-2av_2^2+2av_1^2)+(a^2v_2^2+b^2v_2^2-v_1^2a^2)=0$$
The solutions of this equation are $x_{1/2} =a \pm \frac{v_2 b}{\sqrt{v_1^2-v_2^2}} $ Putting $\frac{dt}{dx}>0$ it is verified for $0<x<a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$ and for $x>a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$.
So the minimum point is for $a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$ and the minimum time is
$$t(a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b (v_1^2+v_2^2)}{v_1v_2 \sqrt{v_1^2-v_2^2}}+\frac{a}{v_1}$$
I'm not sure that I haven't done mistakes, because in the solution on the book it indicates $a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}} $ as the solution of minimum.
and the minimum time
$$t(a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}+\frac{a}{v_1}$$
but, in particular , it distinguishes the case of $\frac{v_2^2}{v_1^2}\ge \frac{a^2}{a^2+b^2}$ in which the minimum time is possible walking from A to B directly.
Your calculations are mostly correct, you just have to interpret the results correctly.
You have a continuous function $t : \Bbb{R} \to \Bbb{R}$ and hence it must attain its minimum on the compact set $[0,a]$ in some point $x_{0} \in [0,a]$.
Moreover, $t$ is differentiable on $\langle 0,a\rangle$ and if it happens that $x_0 \in \langle 0,a\rangle$ then $x_0$ is a local extremum of $t$ so in particular $t'(x_0) = 0$. Assuming $v_1 > v_2$ (if $v_1 \le v_2$, then simply $x_0 = 0$), you correctly found that the possible zeroes of $t'$ are included in $\left\{a- \frac{bv_2}{\sqrt{v_1^2-v_2^2}},a+ \frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right\}$ (as @David K pointed out, the only actual zero of $t'$ is $a- \frac{bv_2}{\sqrt{v_1^2-v_2^2}}$ but it doesn't matter here). Therefore, we conclude that $$x_0 \in \left\{0,a, a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}},a+\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right\}.$$ However, we are interested only in $x_0 \in [0,a]$. Notice that $a+\frac{bv_2}{\sqrt{v_1^2-v_2^2}} > a$ so this option is certainly wrong.
Moreover, we have $$a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}} \in [0,a] \iff \frac{bv_2}{\sqrt{v_1^2-v_2^2}} \le a \iff \frac{v_2}{v_1} \le \frac{a}{\sqrt{a^2+b^2}}.$$
Hence if $\frac{v_2}{v_1} \le \frac{a}{\sqrt{a^2+b^2}}$, then $a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}} \in [0,a]$ and this is indeed the minimum point. Namely,
$$t\left(a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right) = \frac{a}{v_1}+\frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}$$ which is smaller than both $t(0)$ and $t(a)$. We can use Cauchy-Schwarz: $$\frac{a}{v_1}+\frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}\le \sqrt{a^2+b^2}\sqrt{\frac{1}{v_1^2}+\frac{v_1^2-v_2^2}{v_2^2}} = \frac{\sqrt{a^2+b^2}}{v_2} = t(0),$$ $$\frac{a}{v_1}+\frac{b}{v_2} \cdot \underbrace{\frac{\sqrt{v_1^2-v_2^2}}{v_1}}_{\le 1} \le \frac{a}{v_1}+\frac{b}{v_2} = t(a).$$
On the other hand, if $\frac{v_2}{v_1} > \frac{a}{\sqrt{a^2+b^2}}$, then this point is not in $[0,a]$ so it has to be simply $x_0 = 0$ or $x_0 = a$. It is not hard to see that $$t(0) \le t(a) \iff \frac{\sqrt{a^2+b^2}}{v_2} \le \frac{a}{v_1}+\frac{b}{v_2} \iff \frac{v_2}{v_1} \ge \frac{\sqrt{a^2+b^2}-b}a$$ and the latter is true in our case since $$\frac{v_2}{v_1} > \frac{a}{\sqrt{a^2+b^2}} \ge \frac{\sqrt{a^2+b^2}-b}a.$$ So it has to be $x_0 = 0$.
Therefore:
$$x_0 = \begin{cases} a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}, &\text{ if } \frac{v_2}{v_1} \in \left\langle 0, \frac{a}{\sqrt{a^2+b^2}}\right] \\ 0, &\text{ if } \frac{v_2}{v_1} \in \left[\frac{a}{\sqrt{a^2+b^2}}, 1\right\rangle \\ \end{cases}$$
If $v_1 >> v_2$ then $x_0$ approaches $a$, namely $$\lim_{\frac{v_2}{v_1} \to 0} \left(a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right) = \lim_{\frac{v_2}{v_1} \to 0} \left(a-\frac{b}{\sqrt{\left(\frac{v_1}{v_2}\right)^2-1}}\right) = a$$ but it is always faster to go at least somewhat diagonally than straight left and then straight up.
On the other hand, if $v_1$ approaches $v_2$ then you may wonder why $$\lim_{\frac{v_2}{v_1} \to 1} \left(a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right) = -\infty$$ and not $0$ but this expression stops being relevant as soon as $\frac{v_2}{v_1} \ge \frac{a}{\sqrt{a^2+b^2}}$ as is it is faster to simply go diagonally.