For this case we must solve the following quadratic equation:
[tex]x ^ 2 + 4x + 45 = 0[/tex]
Where:
[tex]a = 1\\b = 4\\c = 45[/tex]
The solution will be given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting the values we have:
[tex]x = \frac {-4 \pm \sqrt {4 ^ 2-4 (1) (45)}} {2 (1)}\\x = \frac {-4 \pm \sqrt {16-180}} {2}\\x = \frac {-4 \pm \sqrt {-164}} {2}[/tex]
By definition we have to:
[tex]i ^ 2 = -1[/tex]
So:
[tex]x = \frac {-4 \pm \sqrt {164i ^ 2}} {2}\\x = \frac {-4 \pm i \sqrt {164}} {2}\\x = \frac {-4 \pm i \sqrt {2 ^ 2 * 41}} {2}\\x = \frac {-4 \pm2i \sqrt {41}} {2}\\x = -2 \pm i \sqrt {41}[/tex]
Thus, we have two complex roots:
[tex]x_ {1} = - 2 + i \sqrt {41}\\x_ {2} = - 2-i \sqrt {41}[/tex]
Answer:
[tex]x_ {1} = - 2 + i \sqrt {41}\\x_ {2} = - 2-i \sqrt {41}[/tex]
