To factor the function using the quadratic formula we equate it to zero and solve for x:
[tex]\begin{gathered} 9x^2-149x-234=0 \\ x=\frac{-(-149)\pm\sqrt[]{(-149)^2-4(9)(-234)}}{2(9)} \\ x=\frac{149\pm\sqrt[]{30625}}{18} \\ x=\frac{149\pm175}{18} \\ \text{then} \\ x=\frac{149+175}{18}=18 \\ or \\ x=\frac{149-175}{18}=-\frac{26}{18}=-\frac{13}{9} \end{gathered}[/tex]
Now we write the function as:
[tex]f(x)=(x-a)(x-b)[/tex]
where a and b are the roots we found above, then we have:
[tex]\begin{gathered} f(x)=(x-18)(x-(-\frac{13}{9})) \\ f(x)=(x-18)(9x+13) \end{gathered}[/tex]
Therefore:
[tex]f(x)=(x-18)(9x+13)[/tex]
