We can write the question as a a polynomial as it follows:
[tex]n=n^2-6[/tex]
The values that makethis expression true are the zero of the following polynomial:
[tex]f(n)=n^2-n-6[/tex]
We can use Bhaskara to find the values:
[tex]n_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Where:
[tex]\begin{gathered} a=1 \\ b=-1 \\ c=-6 \end{gathered}[/tex]
Substituting the values, we find:
[tex]n_{1,2}=\frac{1\pm\sqrt[]{1-4(1)(-6)}}{2}=\frac{1\pm\sqrt[]{25}}{2}=\frac{1\pm5}{2}[/tex]
Before solving, we can see that there will realy have two real numbers for n. So the answer is the third optio: Two.
We can go ahead and find the values:
[tex]\begin{gathered} n_1=\frac{1+5}{2}=\frac{6}{2}=3 \\ n_2=\frac{1-5}{2}=-\frac{4}{2}=-2 \end{gathered}[/tex]
