The interval notation is [tex](-\sqrt{11}-4, \sqrt{11}-4)[/tex]
Explanation:
The inequality is [tex]x^{2}+8 x+5<0[/tex]
Let us complete the square, we get,
[tex]x^{2}+8 x+5+4^{2}-4^{2}<0[/tex]
Simplifying, we have,
[tex](x+4)^{2}+5-4^{2}<0[/tex]
[tex](x+4)^{2}-11<0[/tex]
Adding both sides of the equation by 11, we get,
[tex](x+4)^{2}<11[/tex]
Since, we know that for [tex]u^{n}<a[/tex] , if n is even then [tex]-\sqrt[n]{a}<u<\sqrt[n]{a}[/tex]
Thus, writing the above expression in this form, we get,
[tex]-\sqrt{11}<x+4<\sqrt{11}[/tex]
Also, if [tex]a<u<b[/tex] then [tex]a<u[/tex] and [tex]u<b[/tex]
Then ,we have,
[tex]-\sqrt{11}<x+4[/tex] and [tex]x+4<\sqrt{11}[/tex]
Solving the inequalities, we get,
[tex]-\sqrt{11}\ \ \ \ \ \ <x+4\\-\sqrt{11}-4<x[/tex] and [tex]x+4<\sqrt{11}\\x \ \ \ \ \ <\sqrt{11}-4[/tex]
Merging the intervals, we have,
[tex]-\sqrt{11}-4<x<\sqrt{11}-4[/tex]
Hence, writing the solution in interval notation, we have,
[tex](-\sqrt{11}-4, \sqrt{11}-4)[/tex]
Therefore, the answer is [tex](-\sqrt{11}-4, \sqrt{11}-4)[/tex]