For this case we have the following equation:
[tex]2x ^ 2 + x + 2 = 0[/tex]
The roots will be given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}[/tex]
Where:
[tex]a = 2\\b = 1\\c = 2[/tex]
Substituting the values:
[tex]x = \frac {-1 \pm \sqrt {1 ^ 2-4 (2) (2)}} {2 (2)}\\x = \frac {-1 \pm \sqrt {1-16}} {4}\\x = \frac {-1 \pm \sqrt {-15}} {4}\\x = \frac {-1 \pm i \sqrt {15}} {4}[/tex]
We have two roots:
[tex]x_ {1} = \frac {-1 + i \sqrt {15}} {4}\\x_ {2} = \frac {-1-i \sqrt {15}} {4}[/tex]
ANswer:
[tex]x_ {1} = \frac {-1 + i \sqrt {15}} {4}\\x_ {2} = \frac {-1-i \sqrt {15}} {4}[/tex]
