b real!

This is a quadratic equation disguised as a quartic (much harder to solve!).

In order to transform this into a quadratic, let $a = b^2$ .

Then the equation becomes:

$a^2 -11a - 26 = 0$

$\implies (a-13)(a+2) = 0$

This has roots $a=13$ and $a = -2$ .

Since, $b$ is real, the root we car about is the positive one, $a = 13$ , since $a = b^2$ .

So, $b = \pm \sqrt{13}$

And since, $b$ is positive,

$b = | \sqrt{13} | = \boxed{3.6}$ to one decimal place.