I want to solve this equation:
$$t^{10}-t^{9}+ t^{8}- t^{7}+ t^{6}- t^{5}+ t^{4}- t^{3}+ t^{2}- t+1=0$$
with respect to $t$. But I have not a good idea to start. Hence, I am asking about the number of real roots.
Can we deduce the same result for a polynomial of the form:
$$at^{10}-bt^{9}+ ct^{8}-d t^{7}+e t^{6}- ft^{5}+ gt^{4}- ht^{3}+ lt^{2}-m t+r=0$$
where all the coefficient are real and positive.
Let $p(t)$ denote your polynomial. Then it is not hard to see that $$(1+t)p(t)=t^{11}+1,$$ which clearly has $-1$ as its only real root. But $p(-1)=11$, so $p(t)$ has no real roots.
This also shows that over the complex numbers, the roots of $p(t)$ all satisfy $t^{11}=-1$. By Euler's formula we have $$e^{\pi i}=-1,$$ so the roots are all of the form $\exp\left(\tfrac{k}{11}\pi i\right)$ for some integer $k$.
Note that $p(t)$ is in fact the 22nd cyclotomic polynomial; the roots of the $n$-th cyclotomic are the primitive $n$-th roots of unity, which are not real for $n>2$.