We need to find the zeros for the next function:
[tex]f(z)=3z^2−11z−4[/tex]
We can find those zeros using the quadratic formula given by:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Use the form ax²+bx+x.
Where a=3
b= -11
c=-4
Replacing :
[tex]\begin{gathered} x=\frac{-(-11)\pm\sqrt{(-11)^2-4(3)(-4)}}{2(3)} \\ Simplify \\ x=\frac{11\pm\sqrt{169}}{2\ast3} \\ x=\frac{11\pm13}{2\ast3} \\ Therefore: \\ x_1=\frac{11-13}{6}=\frac{-2}{6}=-\frac{1}{3} \\ x_2=\frac{11+13}{6}=\frac{24}{6}=4 \end{gathered}[/tex]
Therefore, the zeros of f (z) are -1/3 and 4.
