# Solving algebraically Quadratic inequalities with modulus abs $(x^2+5x-6) \gt 5$

I tried to solve this inequality algebraically by opening modulus and doing $x^2+5x-6 \gt5$ or $\lt -5$. I got $2$ solution sets with no overlap. When we solve graphically there are $4$ points of intersection. How do we get the answer algebraically ? thanking you in advance Ashwini

We need to solve $$(x^2+5x-6)^2-25>0$$ or $$(x^2+5x-11)(x^2+5x-1)>0$$ or $$\left(x-\frac{-5-\sqrt{69}}{2}\right)\left(x-\frac{-5-\sqrt{29}}{2}\right)\left(x-\frac{-5+\sqrt{29}}{2}\right)\left(x-\frac{-5+\sqrt{69}}{2}\right)>0,$$ which gives the answer: $$\left(-\infty,-\frac{5+\sqrt{69}}{2}\right)\cup\left(-\frac{5+\sqrt{29}}{2},\frac{-5+\sqrt{29}}{2}\right)\cup\left(\frac{-5+\sqrt{69}}{2},+\infty\right).$$