Answer:
[tex]x=\frac{1+5i} {2}[/tex] and [tex]x=\frac{1-5i} {2}[/tex]
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]f(x)=2x^{2} -2x+13[/tex]
Equate the function to zero
[tex]2x^{2} -2x+13=0[/tex]
so
[tex]a=2\\b=-2\\c=13[/tex]
substitute in the formula
[tex]x=\frac{-(-2)\pm\sqrt{-2^{2}-4(2)(13)}} {2(2)}[/tex]
[tex]x=\frac{2\pm\sqrt{-100}} {4}[/tex]
Remember that
[tex]i=\sqrt{-1}[/tex]
so
[tex]x=\frac{2\pm10i} {4}[/tex]
Simplify
[tex]x=\frac{1\pm5i} {2}[/tex]
therefore
[tex]x=\frac{1+5i} {2}[/tex] and [tex]x=\frac{1-5i} {2}[/tex]
