For this case we must solve the following quadratic equation:
[tex]3x ^ 2-2x + 5 = 0[/tex]
Where:
[tex]a = 3\\b = -2\\c = 5[/tex]
The roots are given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting the values we have:
[tex]x = \frac {- (- 2) \pm \sqrt {(- 2) ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {- (- 2) \pm \sqrt {(- 2) ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {2 \pm \sqrt {4-60}} {6}\\x = \frac {2 \pm \sqrt {-56}} {6}[/tex]
By definition we have to:
[tex]i ^ 2 = -1\\x = \frac {2 \pm \sqrt {56i ^ 2}} {6}\\x = \frac {2 \pm i \sqrt {56}} {6}\\x = \frac {2 \pm i \sqrt {2 ^ 2 * 14}} {6}\\x = \frac {2 \pm 2i \sqrt {14}} {6}\\x = \frac {1 \pm i \sqrt {14}} {3}[/tex]
We have two complex roots:
[tex]x_ {1} = \frac {1+ i \sqrt {14}} {3}\\x_ {2} = \frac {1- i \sqrt {14}} {3}[/tex]
Answer:
[tex]x_ {1} = \frac {1+ i \sqrt {14}} {3}\\x_ {2} = \frac {1- i \sqrt {14}} {3}[/tex]
