linear approximation

First, we notice that the domain is $x \ge -1$ .

$1+\frac{x}{2}<\sqrt{1+x}+0.1$

$\implies~ 0.9+0.5x<\sqrt{1+x}$

Since both sides of the inequality are always non-negative over the domain $x \ge -1$ :

$\implies~ (0.9+0.5x)^2<1+x$

$\implies~ 0.81+0.9x+0.25x^2<1+x$

$\implies~ 0.25x^2 - 0.1x - 0.19<0$

Solving this gives $-0.694427<x<1.09443$ which matches the solution you gave.

By the way, the other side doesn't actually give any restriction:

$\sqrt{1+x}-0.1<1+\frac{x}{2}$

$\implies~ \sqrt{1+x}<1.1+0.5x$

$\implies~ 1+x < (1.1+0.5x)^2$

$\implies~ 1+x < 1.21 + 1.1x + 0.25x^2$

$\implies~ 0 < 0.25x^2 + 0.1x + 0.21$

Which is always true, so there is no additional restriction here.