For this case we have the following quadratic equation:
[tex]3x ^ 2 + 13x + 5 = 0[/tex]
Where:
[tex]a = 3\\b = 13\\c = 5[/tex]
Applying the quadratic formula:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting the values:
[tex]x = \frac {-13 \pm \sqrt {13 ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {-13 \pm \sqrt {169-60}} {6}\\x = \frac {-13 \pm \sqrt {109}} {6}[/tex]
Thus, we have two roots:
[tex]x_ {1} = \frac {-13- \sqrt {109}} {6} = - 3.91\\x_ {2} = \frac {-13+ \sqrt {109}} {6} -0.43[/tex]
Answer:
[tex]x_ {1} = - 3.91\\x_ {2} = - 0.43[/tex]
