For this case we have the following quadratic equation:
[tex]3x ^ 2-x + 5 = 0[/tex]
We have to:
[tex]a = 3\\b = -1\\c = 5[/tex]
The quadratic equation states that:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting the values we have:
[tex]x = \frac {- (- 1) \pm \sqrt {(- 1) ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {1 \pm \sqrt {1-60}} {6}\\x = \frac {1 \pm \sqrt {-59}} {6}\\x = \frac {1 \pm \sqrt {59i ^ 2}} {6}\\x = \frac {1 \pm i \sqrt {59}} {6}[/tex]
We have two roots:
[tex]x_ {1} = \frac {1 + i \sqrt {59}} {6}\\x_ {2} = \frac {1-i \sqrt {59}} {6}[/tex]
Answer:
[tex]x_ {1} = \frac {1 + i \sqrt {59}} {6}\\x_ {2} = \frac {1-i \sqrt {59}} {6}[/tex]
