Consider the expression [tex]n^2+16n-64.[/tex]
Since
[tex]D=b^2-4ac=16^2-4\cdot (-64)=256+256=512,\\ \\\sqrt{D}=16\sqrt{2},[/tex]
then
[tex]n_1=\dfrac{-16+16\sqrt{2}}{2}=-8+8\sqrt{2},\ n_2=\dfrac{-16-16\sqrt{2}}{2}=-8-8\sqrt{2}.[/tex]
Therefore, the factored form of [tex]n^2+16n-64[/tex] is [tex](n+8-8\sqrt{2})(n+8+8\sqrt{2}).[/tex]