You neglected that the radicand features 𝑏² rather than 𝑏·|𝑏|, and the former rises as |𝑏| rises whether it’s positive or negative. Hence the equation having no real solutions for 𝑏²<676 implies |𝑏|<26, not 𝑏<26. As a result, 𝑏 ∈ (–26, 26) rather than 𝑏 ∈ (–∞, 26) satisfies the conditions, the smallest integer solution to the question is therefore 𝑏 = –25.
You neglected that the radicand features 𝑏² rather than 𝑏·|𝑏|, and the former rises as |𝑏| rises whether it’s positive or negative. Hence the equation having no real solutions for 𝑏²<676 implies |𝑏|<26, not 𝑏<26. As a result, 𝑏 ∈ (–26, 26) rather than 𝑏 ∈ (–∞, 26) satisfies the conditions, the smallest integer solution to the question is therefore 𝑏 = –25.
I wrote my comment too hastily, you are correct. My comment has been edited, probably while you were typing yours.
You’ll need to edit it again. The solution range is 𝑏 ∈ (–26, 26), not 𝑏 ∈ (–26, 0) but yeah, the smallest integer is still –25.
Also, the last paragraph is weirdly worded:
You don’t need to introduce some “𝑥” of yours (distinct from the quadratic equation’s 𝑥?!) to express –26 < 𝑏 < 26 in words.
Yes, thanks for catching my errors.
The last paragraph still says “less than 0” though