Start with
[tex]1=\dfrac{1}{n^2}+\dfrac{n^2-4n-5}{3n^2}[/tex]
We observe that both fractions are not defined if [tex]n=0[/tex]. So, we will assume [tex]n \neq 0[/tex].
We multiply both numerator and denominator of the first fraction by 3 and we sum the two fractions:
[tex]1=\dfrac{3}{3n^2}+\dfrac{n^2-4n-5}{3n^2} = \dfrac{n^2-4n-2}{3n^2}[/tex]
We multiply both sides by [tex]3n^2[/tex]:
[tex]n^2-4n-2=3n^2[/tex]
We move everything to one side and solve the quadratic equation:
[tex]2n^2+4n+2=0 \iff 2(n^2+2n+1)=0 \iff 2(n+1)^2=0 \iff n+1=0 \iff n=-1[/tex]
We check the solution:
[tex]1=\dfrac{1}{1}+\dfrac{1+4-5}{3} \iff 1 = 1+0[/tex]
which is true
