Answer:
option B
Step-by-step explanation:
[tex]2x^2-4x+16=0[/tex]
We need to solve this equation using quadratic formula
[tex]x= \frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
a=2, b= -4, c=16
Plug in the values in the formula
[tex]x= \frac{-(-4)+-\sqrt{(-4)^2-4(2)(16}}{2(2)}[/tex]
[tex]x= \frac{4+-\sqrt{16-128}}{4}[/tex]
[tex]x= \frac{4+-\sqrt{-112}}{4}[/tex]
Simplify the square root. the value of square root (-1) = 'i'
[tex]x= \frac{4+-4i\sqrt{7}}{4}[/tex]
Now we divide by 4
[tex]x= 1+-i\sqrt{7}[/tex]
So there are two complex roots. since the degree of polynomial is 2
