Answer
See proof below.
Work Step by Step
$g(t)=\displaystyle \frac{1}{1-t}+\sqrt{1+t}-3.1=(1-t)^{-1}+(1+t)^{1/2}-3.1$
is continuous and differentiable on $(-1,1)$.
$g(-0.5)\approx-0.72\lt 0, \qquad g(0.5)\approx 1.12\gt 0$
so the Intermediate Value Theorem guarantees that there is a zero between $-0.5$ and $0.5$.
Check whether $g$ is increasing over $(-1,1)$.
(If it is, the mentioned zero would be the only zero.)
$g'(t)=-(1-t)^{-2}(-1)+\displaystyle \frac{1}{2}(1+t)^{-1/2}=\frac{1}{(1-t)^{2}}+\frac{1}{2\sqrt{1+t}}$
both terms are positive over $(-1,1)$, so
g increases over the interval $(-1,1)$.
So, $g$ has only one zero in $(-1,1)$
