题目根据如表:xxx−3-3−3−2-2−2−1-1−1…\ldots…444555666x2−bx−5x^{2}-bx-5x2−bx−5131313555−1-1−1…\ldots…−1-1−1555131313确定方程x2−bx−5=0x^{2}-bx-5=0x2−bx−5=0的解的取值范围是______.知识点:估算一元二次方程的近似解章节:未标注