To find the values of x such that f(x)=0, substitute that condition into the equation:
[tex]x^2+12x-5=0[/tex]Use the quadratic formula with a=1, b=12 and c=-5.
[tex]\begin{gathered} x=\frac{-(12)\pm\sqrt[]{(12)^2-4(1)(-5)}}{2(1)} \\ \Rightarrow \\ x=\frac{-12\pm\sqrt[]{144+20}}{2} \\ \Rightarrow \\ x=\frac{-12\pm\sqrt[]{164}}{2} \end{gathered}[/tex]Since the number inside the square root is positive, then the solutions are real numbers.
