For this case we must solve a quadratic equation of the form:
[tex]ax ^ 2 + bx + c[/tex]
Which roots are given by:
[tex]\frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}[/tex]
In this case we have:
[tex]x ^ 2 + 3x + 1[/tex]
So:
[tex]a = 1\\b = 3\\c = 1[/tex]
Substituting:
[tex]\frac {-3 \pm \sqrt {3 ^ 2-4 (1) (1)}} {2 (1)} =\\\frac {-3 \pm \sqrt {9-4}} {2} =\\\frac {-3 \pm \sqrt {5}} {2}[/tex]
So, we have two roots:
[tex]x_ {1} = \frac {-3+ \sqrt {5}} {2}\\x_ {2} = \frac {-3- \sqrt {5}} {2}[/tex]
ANswer:
[tex]x_ {1} = \frac {-3+ \sqrt {5}} {2}\\x_ {2} = \frac {-3- \sqrt {5}} {2}[/tex]
