Answer:
[tex]x=0.5(+/-)2.5i[/tex]
Step-by-step explanation:
we know that
The zeros of he function or x-intercepts are the values of x when the function is equal to zero
so
For f(x)=0
we have
[tex]2x^{2} -2x+13=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]2x^{2} -2x+13=0[/tex]
so
[tex]a=2\\b=-2\\c=13[/tex]
substitute in the formula
[tex]x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(2)(13)}} {2(2)}[/tex]
[tex]x=\frac{2(+/-)\sqrt{-100}} {4}[/tex]
Remember that
[tex]i=\sqrt{-1}[/tex]
so
[tex]x=\frac{2(+/-)10i} {4}[/tex]
[tex]x=0.5(+/-)2.5i[/tex]
The function has no real solutions (complex solutions)
That means, the function do not intersect the x-axis
