Since the equation of the hyperbola is
[tex]4x^2-9y^2=36[/tex]Divide all terms by 36
[tex]\begin{gathered} \frac{4x^2}{36}-\frac{9y^2}{36}=\frac{36}{36} \\ \frac{x^2}{9}-\frac{y^2}{4}=1 \end{gathered}[/tex]Since the form of the equation of the hyperbola is
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]By comparing them
[tex]\begin{gathered} a^2=9 \\ b^2=4 \end{gathered}[/tex]Since the foci are (c, 0) and (-c, 0)
Since the value of c can be found from the rule
[tex]c^2=a^2+b^2[/tex]Then
[tex]\begin{gathered} c^2=9+4 \\ c^2=13 \end{gathered}[/tex]Find the square root of both sides
[tex]c=\pm\sqrt[]{13}[/tex]The foci are
[tex](\sqrt[]{13},0),(-\sqrt[]{13},0)[/tex]Then the answer is C
