Answer: Option A
[tex]x=\frac{1+\sqrt{83}i}{6}[/tex] or [tex]x=\frac{1-\sqrt{83}i}{6}[/tex]
Step-by-step explanation:
Use the quadratic formula to find the zeros of the function.
For a function of the form
[tex]ax ^ 2 + bx + c = 0[/tex]
The quadratic formula is:
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
In this case the function is:
[tex]3x^2-x+7=0[/tex]
So
[tex]a=3\\b=-1\\c=7[/tex]
Then using the quadratic formula we have that:
[tex]x=\frac{-(-1)\±\sqrt{(-1)^2-4(3)(7)}}{2(3)}[/tex]
[tex]x=\frac{1\±\sqrt{1-84}}{6}[/tex]
[tex]x=\frac{1\±\sqrt{-83}}{6}[/tex]
Remember that [tex]\sqrt{-1}=i[/tex]
[tex]x=\frac{1\±\sqrt{83}*\sqrt{-1}}{6}[/tex]
[tex]x=\frac{1\±\sqrt{83}i}{6}[/tex]
[tex]x=\frac{1+\sqrt{83}i}{6}[/tex] or [tex]x=\frac{1-\sqrt{83}i}{6}[/tex]
