For this case we have the following quadratic equation:
[tex]3x ^ 2 + 4x-6 = 0[/tex]
Where:
[tex]a = 3\\b = 4\\c = -6[/tex]
We solve using the quadratic formula:
[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 (3) (- 6)}} {2 (3)}\\x = \frac {-4 \pm \sqrt {16 + 72}} {6}\\x = \frac {-4 \pm \sqrt {88}} {6}\\x = \frac {-4 \pm \sqrt {2 ^ 2 * 22}} {6}\\x = \frac {-4 \pm2 \sqrt {22}} {6}\\x = \frac {-2 \pm \sqrt {22}} {3}[/tex]
We have two roots:
[tex]x_ {1} = \frac {-2+ \sqrt {22}} {3}\\x_ {2} = \frac {-2- \sqrt {22}} {3}[/tex]
Answer:
[tex]x_ {1} = \frac {-2+ \sqrt {22}}{3}\\x_ {2} = \frac {-2- \sqrt {22}} {3}[/tex]
