This is a continuous function. Plugging in 1 we get 41+C. Plugging in -1 we get the same, 41+C. Plugging in 0 we get C.
Choose C < 0 and C > -41. Then by the intermediate value theorem of Calculus, as the function goes from 41+C at x=-1 to C at 0 it crosses the x-axis, and it crosses it again as it goes from 0 to 41+C at x=1. That gives (at least) two real roots.
If you mean plural "values", i.e. you want to characterize all of the values of C that work, then certainly C > 0 doesn't work because all the other summands are perfect squares, hence are non-negative, so adding them gives something >= 0, and then adding C will take us above 0. So C <= 0 is an upper bound. We show that any value of C < 0 works. Let C be < 0. Then choose x to be the sixth (positive) root of -C. For example, if C = -729 then take x to be 3. We conclude
x6+9x4+26x2+C > x6 + C >= 0, and likewise taking x to be the negative square root of -C, so the intermediate value theorem argument works again, giving at least two roots.
The function has one real root if C = 0.
the function will have one or more real roots if and only if C <= 0
the function will have exactly one real root if and only if C = 0
It is also easy to show that the function has zero, one, or exactly two real roots by considering the first derivative, which is positive to the right of 0 and negative to the left of 0, hence there are no local extrema other than the minimum at 0.